A table that is partitioned by range is partitioned in such a way that each partition contains rows for which the partitioning expression value lies within a given range. Ranges should be contiguous but not overlapping, and are defined using the VALUES LESS THAN operator. For the next few examples, suppose that you are creating a table such as the following to hold personnel records for a chain of 20 video stores, numbered 1 through 20:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT NOT NULL,
    store_id INT NOT NULL
);

This table can be partitioned by range in a number of ways, depending on your needs. One way would be to use the store_id column. For instance, you might decide to partition the table 4 ways by adding a PARTITION BY RANGE clause as shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT NOT NULL,
    store_id INT NOT NULL
)
PARTITION BY RANGE (store_id) (
    PARTITION p0 VALUES LESS THAN (6),
    PARTITION p1 VALUES LESS THAN (11),
    PARTITION p2 VALUES LESS THAN (16),
    PARTITION p3 VALUES LESS THAN (21)
);

In this partitioning scheme, all rows corresponding to employees working at stores 1 through 5 are stored in partition p0, to those employed at stores 6 through 10 are stored in partition p1, and so on. Note that each partition is defined in order, from lowest to highest. This is a requirement of the PARTITION BY RANGE syntax; you can think of it as being analogous to a series of if ... elseif ... statements in C or Java in this regard.

It is easy to determine that a new row containing the data (72, 'Michael', 'Widenius', '1998-06-25', NULL, 13) is inserted into partition p2, but what happens when your chain adds a 21^st store? Under this scheme, there is no rule that covers a row whose store_id is greater than 20, so an error results because the server does not know where to place it. You can keep this from occurring by using a “catchall” VALUES LESS THAN clause in the CREATE TABLE statement that provides for all values greater than the highest value explicitly named:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT NOT NULL,
    store_id INT NOT NULL
)
PARTITION BY RANGE (store_id) (
    PARTITION p0 VALUES LESS THAN (6),
    PARTITION p1 VALUES LESS THAN (11),
    PARTITION p2 VALUES LESS THAN (16),
    PARTITION p3 VALUES LESS THAN MAXVALUE
);

MAXVALUE represents an integer value that is always greater than the largest possible integer value (in mathematical language, it serves as a least upper bound). Now, any rows whose store_id column value is greater than or equal to 16 (the highest value defined) are stored in partition p3. At some point in the future — when the number of stores has increased to 25, 30, or more — you can use an ALTER TABLE statement to add new partitions for stores 21-25, 26-30, and so on (see Section 17.3, “Partition Management”, for details of how to do this).

In much the same fashion, you could partition the table based on employee job codes — that is, based on ranges of job_code column values. For example — assuming that two-digit job codes are used for regular (in-store) workers, three-digit codes are used for office and support personnel, and four-digit codes are used for management positions — you could create this partitioned table using the following statement:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT NOT NULL,
    store_id INT NOT NULL
)
PARTITION BY RANGE (job_code) (
    PARTITION p0 VALUES LESS THAN (100),
    PARTITION p1 VALUES LESS THAN (1000),
    PARTITION p2 VALUES LESS THAN (10000)
);

In this instance, all rows relating to in-store workers would be stored in partition p0, those relating to office and support staff in p1, and those relating to managers in partition p2.

It is also possible to use an expression in VALUES LESS THAN clauses. However, MySQL must be able to evaluate the expression's return value as part of a LESS THAN (<) comparison.

Rather than splitting up the table data according to store number, you can use an expression based on one of the two DATE columns instead. For example, let us suppose that you wish to partition based on the year that each employee left the company; that is, the value of YEAR(separated). An example of a CREATE TABLE statement that implements such a partitioning scheme is shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
)
PARTITION BY RANGE ( YEAR(separated) ) (
    PARTITION p0 VALUES LESS THAN (1991),
    PARTITION p1 VALUES LESS THAN (1996),
    PARTITION p2 VALUES LESS THAN (2001),
    PARTITION p3 VALUES LESS THAN MAXVALUE
);

In this scheme, for all employees who left before 1991, the rows are stored in partition p0; for those who left in the years 1991 through 1995, in p1; for those who left in the years 1996 through 2000, in p2; and for any workers who left after the year 2000, in p3.

Range partitioning is particularly useful when:

A variant on this type of partitioning, RANGE COLUMNS partitioning, was introduced in MySQL 5.5.0. Partitioning by RANGE COLUMNS makes it possible to employ multiple columns for defining partitioning ranges that apply both to placement of rows in partitions and for determining the inclusion or exclusion of specific partitions when performing partition pruning. See Section 17.2.3.1, “Range columns partitioning”, for more information.

If you wish to use partitioning based on ranges or intervals of time in MySQL 5.5, you have two options:

List partitioning in MySQL is similar to range partitioning in many ways. As in partitioning by RANGE, each partition must be explicitly defined. The chief difference is that, in list partitioning, each partition is defined and selected based on the membership of a column value in one of a set of value lists, rather than in one of a set of contiguous ranges of values. This is done by using PARTITION BY LIST(expr) where expr is a column value or an expression based on a column value and returning an integer value, and then defining each partition by means of a VALUES IN (value_list), where value_list is a comma-separated list of integers.

Unlike the case with partitions defined by range, list partitions do not need to be declared in any particular order. For more detailed syntactical information, see Section 12.1.14, “CREATE TABLE Syntax”.

For the examples that follow, we assume that the basic definition of the table to be partitioned is provided by the CREATE TABLE statement shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
);

(This is the same table used as a basis for the examples in Section 17.2.1, “RANGE Partitioning”.)

Suppose that there are 20 video stores distributed among 4 franchises as shown in the following table.

To partition this table in such a way that rows for stores belonging to the same region are stored in the same partition, you could use the CREATE TABLE statement shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
)
PARTITION BY LIST(store_id) (
    PARTITION pNorth VALUES IN (3,5,6,9,17),
    PARTITION pEast VALUES IN (1,2,10,11,19,20),
    PARTITION pWest VALUES IN (4,12,13,14,18),
    PARTITION pCentral VALUES IN (7,8,15,16)
);

This makes it easy to add or drop employee records relating to specific regions to or from the table. For instance, suppose that all stores in the West region are sold to another company. Beginning with MySQL 5.5.0, all rows relating to employees working at stores in that region can be deleted with the query ALTER TABLE employees TRUNCATE PARTITION pWest, which can be executed much more efficiently than the equivalent DELETE statement DELETE FROM employees WHERE store_id IN (4,12,13,14,18);. (Using ALTER TABLE employees DROP PARTITION pWest would also delete all of these rows, but would also remove the partition pWest from the definition of the table; you would need to use an ALTER TABLE ... ADD PARTITION statement to restore the table's original partitioning scheme.)

As with RANGE partitioning, it is possible to combine LIST partitioning with partitioning by hash or key to produce a composite partitioning (subpartitioning). See Section 17.2.6, “Subpartitioning”.

Unlike the case with RANGE partitioning, there is no “catch-all” such as MAXVALUE; all expected values for the partitioning expression should be covered in PARTITION ... VALUES IN (...) clauses. An INSERT statement containing an unmatched partitioning column value fails with an error, as shown in this example:

mysql> CREATE TABLE h2 (
    ->   c1 INT,
    ->   c2 INT
    -> )
    -> PARTITION BY LIST(c1) (
    ->   PARTITION p0 VALUES IN (1, 4, 7),
    ->   PARTITION p1 VALUES IN (2, 5, 8)
    -> );
Query OK, 0 rows affected (0.11 sec)

mysql> INSERT INTO h2 VALUES (3, 5);
ERROR 1525 (HY000): Table has no partition for value 3

When inserting multiple rows using a single INSERT statement, any rows coming before the row containing the unmatched value are inserted, but any coming after it are not:

mysql> SELECT * FROM h2;
Empty set (0.00 sec)

mysql> INSERT INTO h2 VALUES (4, 7), (3, 5), (6, 0);
ERROR 1525 (HY000): Table has no partition for value 3
mysql> SELECT * FROM h2;
+------+------+
| c1   | c2   |
+------+------+
|    4 |    7 |
+------+------+
1 row in set (0.00 sec)

You can cause this type of error to be ignored by using the IGNORE key word. If you do so, rows containing unmatched partitioning column values are not inserted, but any rows with matching values are inserted, and no errors are reported:

mysql> TRUNCATE h2;
Query OK, 1 row affected (0.00 sec)

mysql> SELECT * FROM h2;
Empty set (0.00 sec)

mysql> INSERT IGNORE INTO h2 VALUES (2, 5), (6, 10), (7, 5), (3, 1), (1, 9);
Query OK, 3 rows affected (0.00 sec)
Records: 5  Duplicates: 2  Warnings: 0

mysql> SELECT * FROM h2;
+------+------+
| c1   | c2   |
+------+------+
|    7 |    5 |
|    1 |    9 |
|    2 |    5 |
+------+------+
3 rows in set (0.00 sec)

Beginning with MySQL 5.5.0, MySQL provides support for LIST COLUMNS partitioning. This is a variant of LIST partitioning that allows you to use columns of types other than integer types for partitioning columns, as well as to use multiple columns as partitioning keys. For more information, see Section 17.2.3.2, “List columns partitioning”.

The next two sections discuss COLUMNS partitioning, which are variants on RANGE and LIST partitioning that were introduced in MySQL 5.5.0. COLUMNS partitioning allows the use of multiple columns in partitioning keys. The multiple columns are taken into account both for the purpose of placing rows in partitions and for the determination of which partitions are to be checked for matching rows in partition pruning.

In addition, both RANGE COLUMNS partitioning and LIST COLUMNS partitioning support the use of non-integer columns for defining value ranges or list members. The permitted data types are shown in the following list:

The discussions of RANGE COLUMNS and LIST COLUMNS partitioning in the next two sections assume that you are already familiar with partitioning based on ranges and lists as supported in MySQL 5.1 and later; for more information about these, see Section 17.2.1, “RANGE Partitioning”, and Section 17.2.2, “LIST Partitioning”, respectively.

CREATE TABLE table_name
PARTITIONED BY RANGE COLUMNS(column_list) (
    PARTITION partition_name VALUES LESS THAN (value_list)[,
    PARTITION partition_name VALUES LESS THAN (value_list)][,
    ...]
) 

column_list:
    column_name[, column_name][, ...]

value_list:
    value[, value][, ...]

mysql> CREATE TABLE rcx (
    ->     a INT,
    ->     b INT,
    ->     c CHAR(3),
    ->     d INT
    -> )
    -> PARTITION BY RANGE COLUMNS(a,d,c) (
    ->     PARTITION p0 VALUES LESS THAN (5,10,'ggg'),
    ->     PARTITION p1 VALUES LESS THAN (10,20,'mmmm'),
    ->     PARTITION p2 VALUES LESS THAN (15,30,'sss'),
    ->     PARTITION p3 VALUES LESS THAN (MAXVALUE,MAXVALUE,MAXVALUE)
    -> );
Query OK, 0 rows affected (0.15 sec)

CREATE TABLE r1 (
    a INT,
    b INT
)
PARTITION BY RANGE (a)  (
    PARTITION p0 VALUES LESS THAN (5),
    PARTITION p1 VALUES LESS THAN (MAXVALUE)
);

mysql> INSERT INTO r1 VALUES (5,10), (5,11), (5,12);
Query OK, 3 rows affected (0.00 sec)
Records: 3  Duplicates: 0  Warnings: 0

mysql> SELECT PARTITION_NAME,TABLE_ROWS
    ->     FROM INFORMATION_SCHEMA.PARTITIONS
    ->     WHERE TABLE_NAME = 'r1';
+----------------+------------+
| PARTITION_NAME | TABLE_ROWS |
+----------------+------------+
| p0             |          0 |
| p1             |          3 |
+----------------+------------+
2 rows in set (0.00 sec)

CREATE TABLE rc1 (
    a INT, 
    b INT
) 
PARTITION BY RANGE COLUMNS(a, b) (
    PARTITION p0 VALUES LESS THAN (5, 12),
    PARTITION p3 VALUES LESS THAN (MAXVALUE, MAXVALUE)
);

mysql> INSERT INTO rc1 VALUES (5,10), (5,11), (5,12);
Query OK, 3 rows affected (0.00 sec)
Records: 3  Duplicates: 0  Warnings: 0

mysql> SELECT PARTITION_NAME,TABLE_ROWS
    ->     FROM INFORMATION_SCHEMA.PARTITIONS
    ->     WHERE TABLE_NAME = 'rc1';
+--------------+----------------+------------+
| TABLE_SCHEMA | PARTITION_NAME | TABLE_ROWS |
+--------------+----------------+------------+
| p            | p0             |          2 |
| p            | p1             |          1 |
+--------------+----------------+------------+
2 rows in set (0.00 sec)

mysql> SELECT (5,10) < (5,12), (5,11) < (5,12), (5,12) < (5,12);
+-----------------+-----------------+-----------------+
| (5,10) < (5,12) | (5,11) < (5,12) | (5,12) < (5,12) |
+-----------------+-----------------+-----------------+
|               1 |               1 |               0 |
+-----------------+-----------------+-----------------+
1 row in set (0.00 sec)

SELECT ROW(5,10) < ROW(5,12), ROW(5,11) < ROW(5,12), ROW(5,12) < ROW(5,12);

CREATE TABLE rx (
    a INT,
    b INT
)
PARTITION BY RANGE COLUMNS (a)  (
    PARTITION p0 VALUES LESS THAN (5),
    PARTITION p1 VALUES LESS THAN (MAXVALUE)
); 

mysql> INSERT INTO rx VALUES (5,10), (5,11), (5,12);
Query OK, 3 rows affected (0.00 sec)
Records: 3  Duplicates: 0  Warnings: 0

mysql> SELECT PARTITION_NAME,TABLE_ROWS
    ->     FROM INFORMATION_SCHEMA.PARTITIONS
    ->     WHERE TABLE_NAME = 'rx';
+--------------+----------------+------------+
| TABLE_SCHEMA | PARTITION_NAME | TABLE_ROWS |
+--------------+----------------+------------+
| p            | p0             |          0 |
| p            | p1             |          3 |
+--------------+----------------+------------+
2 rows in set (0.00 sec)

CREATE TABLE rc2 (
    a INT,
    b INT
)
PARTITION BY RANGE COLUMNS(a,b) (
    PARTITION p0 VALUES LESS THAN (0,10),
    PARTITION p1 VALUES LESS THAN (10,20),
    PARTITION p2 VALUES LESS THAN (10,30),
    PARTITION p3 VALUES LESS THAN (MAXVALUE,MAXVALUE)
 );
 
CREATE TABLE rc3 (
    a INT,
    b INT
)
PARTITION BY RANGE COLUMNS(a,b) (
    PARTITION p0 VALUES LESS THAN (0,10),
    PARTITION p1 VALUES LESS THAN (10,20),
    PARTITION p2 VALUES LESS THAN (10,30),
    PARTITION p3 VALUES LESS THAN (10,35),
    PARTITION p4 VALUES LESS THAN (20,40),
    PARTITION p5 VALUES LESS THAN (MAXVALUE,MAXVALUE)
 );

CREATE TABLE rc4 (
    a INT,
    b INT,
    c INT
)
PARTITION BY RANGE COLUMNS(a,b,c) (
    PARTITION p0 VALUES LESS THAN (0,25,50),
    PARTITION p1 VALUES LESS THAN (10,20,100),
    PARTITION p2 VALUES LESS THAN (10,30,50)
    PARTITION p3 VALUES LESS THAN (MAXVALUE,MAXVALUE,MAXVALUE) 
 );

mysql> SELECT (0,25,50) < (10,20,100), (10,20,100) < (10,30,50);
+-------------------------+--------------------------+
| (0,25,50) < (10,20,100) | (10,20,100) < (10,30,50) |
+-------------------------+--------------------------+
|                       1 |                        1 |
+-------------------------+--------------------------+
1 row in set (0.00 sec)

mysql> CREATE TABLE rcf (
    ->     a INT,
    ->     b INT,
    ->     c INT
    -> )
    -> PARTITION BY RANGE COLUMNS(a,b,c) (
    ->     PARTITION p0 VALUES LESS THAN (0,25,50),
    ->     PARTITION p1 VALUES LESS THAN (20,20,100),
    ->     PARTITION p2 VALUES LESS THAN (10,30,50),
    ->     PARTITION p3 VALUES LESS THAN (MAXVALUE,MAXVALUE,MAXVALUE)
    ->  );
ERROR 1493 (HY000): VALUES LESS THAN value must be strictly increasing for each partition

mysql> SELECT (0,25,50) < (20,20,100), (20,20,100) < (10,30,50);
+-------------------------+--------------------------+
| (0,25,50) < (20,20,100) | (20,20,100) < (10,30,50) |
+-------------------------+--------------------------+
|                       1 |                        0 |
+-------------------------+--------------------------+
1 row in set (0.00 sec)

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT NOT NULL,
    store_id INT NOT NULL
);

CREATE TABLE employees_by_lname (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT NOT NULL,
    store_id INT NOT NULL
)
PARTITION BY RANGE COLUMNS (lname)  (
    PARTITION p0 VALUES LESS THAN ('g'),
    PARTITION p1 VALUES LESS THAN ('m'),
    PARTITION p2 VALUES LESS THAN ('t'),
    PARTITION p3 VALUES LESS THAN (MAXVALUE)
);

ALTER TABLE employees PARTITION BY RANGE COLUMNS (lname)  (
    PARTITION p0 VALUES LESS THAN ('g'),
    PARTITION p1 VALUES LESS THAN ('m'),
    PARTITION p2 VALUES LESS THAN ('t'),
    PARTITION p3 VALUES LESS THAN (MAXVALUE)
);

ALTER TABLE employees PARTITION BY RANGE COLUMNS (hired)  (
    PARTITION p0 VALUES LESS THAN ('1970-01-01'),
    PARTITION p1 VALUES LESS THAN ('1980-01-01'),
    PARTITION p2 VALUES LESS THAN ('1990-01-01'),
    PARTITION p3 VALUES LESS THAN ('2000-01-01'),
    PARTITION p4 VALUES LESS THAN ('2010-01-01'),
    PARTITION p5 VALUES LESS THAN (MAXVALUE)
);

CREATE TABLE customers_1 (
    first_name VARCHAR(25),
    last_name VARCHAR(25),
    street_1 VARCHAR(30),
    street_2 VARCHAR(30),
    city VARCHAR(15),
    renewal DATE
)
PARTITION BY LIST COLUMNS(city) (
    PARTITION pRegion_1 VALUES IN('Oskarshamn', 'Högsby', 'Mönsterås'),
    PARTITION pRegion_2 VALUES IN('Vimmerby', 'Hultsfred', 'Västervik'),
    PARTITION pRegion_3 VALUES IN('Nässjö', 'Eksjö', 'Vetlanda'),
    PARTITION pRegion_4 VALUES IN('Uppvidinge', 'Alvesta', 'Växjo')
);

CREATE TABLE customers_2 (
    first_name VARCHAR(25),
    last_name VARCHAR(25),
    street_1 VARCHAR(30),
    street_2 VARCHAR(30),
    city VARCHAR(15),
    renewal DATE
)
PARTITION BY LIST COLUMNS(renewal) (
    PARTITION pWeek_1 VALUES IN('2010-02-01', '2010-02-02', '2010-02-03',
        '2010-02-04', '2010-02-05', '2010-02-06', '2010-02-07'),
    PARTITION pWeek_2 VALUES IN('2010-02-08', '2010-02-09', '2010-02-10',
        '2010-02-11', '2010-02-12', '2010-02-13', '2010-02-14'),
    PARTITION pWeek_3 VALUES IN('2010-02-15', '2010-02-16', '2010-02-17',
        '2010-02-18', '2010-02-19', '2010-02-20', '2010-02-21'),
    PARTITION pWeek_4 VALUES IN('2010-02-22', '2010-02-23', '2010-02-24',
        '2010-02-25', '2010-02-26', '2010-02-27', '2010-02-28')
);

CREATE TABLE customers_3 (
    first_name VARCHAR(25),
    last_name VARCHAR(25),
    street_1 VARCHAR(30),
    street_2 VARCHAR(30),
    city VARCHAR(15),
    renewal DATE
)
PARTITION BY RANGE COLUMNS(renewal) (
    PARTITION pWeek_1 VALUES LESS THAN('2010-02-09'),
    PARTITION pWeek_2 VALUES LESS THAN('2010-02-15'),
    PARTITION pWeek_3 VALUES LESS THAN('2010-02-22'),
    PARTITION pWeek_4 VALUES LESS THAN('2010-03-01')
);

Partitioning by HASH is used primarily to ensure an even distribution of data among a predetermined number of partitions. With range or list partitioning, you must specify explicitly into which partition a given column value or set of column values is to be stored; with hash partitioning, MySQL takes care of this for you, and you need only specify a column value or expression based on a column value to be hashed and the number of partitions into which the partitioned table is to be divided.

To partition a table using HASH partitioning, it is necessary to append to the CREATE TABLE statement a PARTITION BY HASH (expr) clause, where expr is an expression that returns an integer. This can simply be the name of a column whose type is one of MySQL's integer types. In addition, you will most likely want to follow this with a PARTITIONS num clause, where num is a positive integer representing the number of partitions into which the table is to be divided.

For example, the following statement creates a table that uses hashing on the store_id column and is divided into 4 partitions:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
)
PARTITION BY HASH(store_id)
PARTITIONS 4;

If you do not include a PARTITIONS clause, the number of partitions defaults to 1.

Using the PARTITIONS keyword without a number following it results in a syntax error.

You can also use an SQL expression that returns an integer for expr. For instance, you might want to partition based on the year in which an employee was hired. This can be done as shown here:

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
)
PARTITION BY HASH( YEAR(hired) )
PARTITIONS 4;

expr must return a nonconstant, nonrandom integer value (in other words, it should be varying but deterministic), and must not contain any prohibited constructs as described in Section 17.5, “Restrictions and Limitations on Partitioning”. You should also keep in mind that this expression is evaluated each time a row is inserted or updated (or possibly deleted); this means that very complex expressions may give rise to performance issues, particularly when performing operations (such as batch inserts) that affect a great many rows at one time.

The most efficient hashing function is one which operates upon a single table column and whose value increases or decreases consistently with the column value, as this allows for “pruning” on ranges of partitions. That is, the more closely that the expression varies with the value of the column on which it is based, the more efficiently MySQL can use the expression for hash partitioning.

For example, where date_col is a column of type DATE, then the expression TO_DAYS(date_col) is said to vary directly with the value of date_col, because for every change in the value of date_col, the value of the expression changes in a consistent manner. The variance of the expression YEAR(date_col) with respect to date_col is not quite as direct as that of TO_DAYS(date_col), because not every possible change in date_col produces an equivalent change in YEAR(date_col). Even so, YEAR(date_col) is a good candidate for a hashing function, because it varies directly with a portion of date_col and there is no possible change in date_col that produces a disproportionate change in YEAR(date_col).

By way of contrast, suppose that you have a column named int_col whose type is INT. Now consider the expression POW(5-int_col,3) + 6. This would be a poor choice for a hashing function because a change in the value of int_col is not guaranteed to produce a proportional change in the value of the expression. Changing the value of int_col by a given amount can produce by widely different changes in the value of the expression. For example, changing int_col from 5 to 6 produces a change of -1 in the value of the expression, but changing the value of int_col from 6 to 7 produces a change of -7 in the expression value.

In other words, the more closely the graph of the column value versus the value of the expression follows a straight line as traced by the equation y=nx where n is some nonzero constant, the better the expression is suited to hashing. This has to do with the fact that the more nonlinear an expression is, the more uneven the distribution of data among the partitions it tends to produce.

In theory, pruning is also possible for expressions involving more than one column value, but determining which of such expressions are suitable can be quite difficult and time-consuming. For this reason, the use of hashing expressions involving multiple columns is not particularly recommended.

When PARTITION BY HASH is used, MySQL determines which partition of num partitions to use based on the modulus of the result of the user function. In other words, for an expression expr, the partition in which the record is stored is partition number N, where N = MOD(expr, num). Suppose that table t1 is defined as follows, so that it has 4 partitions:

CREATE TABLE t1 (col1 INT, col2 CHAR(5), col3 DATE)
    PARTITION BY HASH( YEAR(col3) )
    PARTITIONS 4;

If you insert a record into t1 whose col3 value is '2005-09-15', then the partition in which it is stored is determined as follows:

MOD(YEAR('2005-09-01'),4)
=  MOD(2005,4)
=  1

MySQL 5.5 also supports a variant of HASH partitioning known as linear hashing which employs a more complex algorithm for determining the placement of new rows inserted into the partitioned table. See Section 17.2.4.1, “LINEAR HASH Partitioning”, for a description of this algorithm.

The user function is evaluated each time a record is inserted or updated. It may also — depending on the circumstances — be evaluated when records are deleted.

CREATE TABLE employees (
    id INT NOT NULL,
    fname VARCHAR(30),
    lname VARCHAR(30),
    hired DATE NOT NULL DEFAULT '1970-01-01',
    separated DATE NOT NULL DEFAULT '9999-12-31',
    job_code INT,
    store_id INT
)
PARTITION BY LINEAR HASH( YEAR(hired) )
PARTITIONS 4;

CREATE TABLE t1 (col1 INT, col2 CHAR(5), col3 DATE)
    PARTITION BY LINEAR HASH( YEAR(col3) )
    PARTITIONS 6;

V = POWER(2, CEILING( LOG(2,6) )) = 8
N = YEAR('2003-04-14') & (8 - 1)
   = 2003 & 7
   = 3

(3 >= 6 is FALSE: record stored in partition #3)

V = 8
N = YEAR('1998-10-19') & (8-1)
  = 1998 & 7
  = 6

(6 >= 6 is TRUE: additional step required)

N = 6 & CEILING(8 / 2)
  = 6 & 3
  = 2

(2 >= 6 is FALSE: record stored in partition #2)

Partitioning by key is similar to partitioning by hash, except that where hash partitioning employs a user-defined expression, the hashing function for key partitioning is supplied by the MySQL server. MySQL Cluster uses MD5() for this purpose; for tables using other storage engines, the server employs its own internal hashing function which is based on the same algorithm as PASSWORD().

The syntax rules for CREATE TABLE ... PARTITION BY KEY are similar to those for creating a table that is partitioned by hash. The major differences are that:

It is also possible to partition a table by linear key. Here is a simple example:

CREATE TABLE tk (
    col1 INT NOT NULL,
    col2 CHAR(5),
    col3 DATE
)
PARTITION BY LINEAR KEY (col1)
PARTITIONS 3;

Using LINEAR has the same effect on KEY partitioning as it does on HASH partitioning, with the partition number being derived using a powers-of-two algorithm rather than modulo arithmetic. See Section 17.2.4.1, “LINEAR HASH Partitioning”, for a description of this algorithm and its implications.

Subpartitioning — also known as composite partitioning — is the further division of each partition in a partitioned table. For example, consider the following CREATE TABLE statement:

CREATE TABLE ts (id INT, purchased DATE)
    PARTITION BY RANGE( YEAR(purchased) )
    SUBPARTITION BY HASH( TO_DAYS(purchased) )
    SUBPARTITIONS 2 (
        PARTITION p0 VALUES LESS THAN (1990),
        PARTITION p1 VALUES LESS THAN (2000),
        PARTITION p2 VALUES LESS THAN MAXVALUE
    );

Table ts has 3 RANGE partitions. Each of these partitions — p0, p1, and p2 — is further divided into 2 subpartitions. In effect, the entire table is divided into 3 * 2 = 6 partitions. However, due to the action of the PARTITION BY RANGE clause, the first 2 of these store only those records with a value less than 1990 in the purchased column.

In MySQL 5.5, it is possible to subpartition tables that are partitioned by RANGE or LIST. Subpartitions may use either HASH or KEY partitioning. This is also known as composite partitioning.

It is also possible to define subpartitions explicitly using SUBPARTITION clauses to specify options for individual subpartitions. For example, a more verbose fashion of creating the same table ts as shown in the previous example would be:

CREATE TABLE ts (id INT, purchased DATE)
    PARTITION BY RANGE( YEAR(purchased) )
    SUBPARTITION BY HASH( TO_DAYS(purchased) ) (
        PARTITION p0 VALUES LESS THAN (1990) (
            SUBPARTITION s0,
            SUBPARTITION s1
        ),
        PARTITION p1 VALUES LESS THAN (2000) (
            SUBPARTITION s2,
            SUBPARTITION s3
        ),
        PARTITION p2 VALUES LESS THAN MAXVALUE (
            SUBPARTITION s4,
            SUBPARTITION s5
        )
    );

Some syntactical items of note:

Subpartitions can be used with especially large tables to distribute data and indexes across many disks. Suppose that you have 6 disks mounted as /disk0, /disk1, /disk2, and so on. Now consider the following example:

CREATE TABLE ts (id INT, purchased DATE)
    PARTITION BY RANGE( YEAR(purchased) )
    SUBPARTITION BY HASH( TO_DAYS(purchased) ) (
        PARTITION p0 VALUES LESS THAN (1990) (
            SUBPARTITION s0
                DATA DIRECTORY = '/disk0/data'
                INDEX DIRECTORY = '/disk0/idx',
            SUBPARTITION s1
                DATA DIRECTORY = '/disk1/data'
                INDEX DIRECTORY = '/disk1/idx'
        ),
        PARTITION p1 VALUES LESS THAN (2000) (
            SUBPARTITION s2
                DATA DIRECTORY = '/disk2/data'
                INDEX DIRECTORY = '/disk2/idx',
            SUBPARTITION s3
                DATA DIRECTORY = '/disk3/data'
                INDEX DIRECTORY = '/disk3/idx'
        ),
        PARTITION p2 VALUES LESS THAN MAXVALUE (
            SUBPARTITION s4
                DATA DIRECTORY = '/disk4/data'
                INDEX DIRECTORY = '/disk4/idx',
            SUBPARTITION s5
                DATA DIRECTORY = '/disk5/data'
                INDEX DIRECTORY = '/disk5/idx'
        )
    );

In this case, a separate disk is used for the data and for the indexes of each RANGE. Many other variations are possible; another example might be:

CREATE TABLE ts (id INT, purchased DATE)
    PARTITION BY RANGE(YEAR(purchased))
    SUBPARTITION BY HASH( TO_DAYS(purchased) ) (
        PARTITION p0 VALUES LESS THAN (1990) (
            SUBPARTITION s0a
                DATA DIRECTORY = '/disk0'
                INDEX DIRECTORY = '/disk1',
            SUBPARTITION s0b
                DATA DIRECTORY = '/disk2'
                INDEX DIRECTORY = '/disk3'
        ),
        PARTITION p1 VALUES LESS THAN (2000) (
            SUBPARTITION s1a
                DATA DIRECTORY = '/disk4/data'
                INDEX DIRECTORY = '/disk4/idx',
            SUBPARTITION s1b
                DATA DIRECTORY = '/disk5/data'
                INDEX DIRECTORY = '/disk5/idx'
        ),
        PARTITION p2 VALUES LESS THAN MAXVALUE (
            SUBPARTITION s2a,
            SUBPARTITION s2b
        )
    );

Here, the storage is as follows:

The DATA DIRECTORY and INDEX DIRECTORY options are disallowed when the NO_DIR_IN_CREATE server SQL mode is in effect. This is true for partitions and subpartitions.

Partitioning in MySQL does nothing to disallow NULL as the value of a partitioning expression, whether it is a column value or the value of a user-supplied expression. Even though it is permitted to use NULL as the value of an expression that must otherwise yield an integer, it is important to keep in mind that NULL is not a number. The partitioning implementation treats NULL as being less than any non-NULL value, just as ORDER BY does.

This means that treatment of NULL varies between partitioning of different types, and may produce behavior which you do not expect if you are not prepared for it. This being the case, we discuss in this section how each MySQL partitioning type handles NULL values when determining the partition in which a row should be stored, and provide examples for each.

Handling of NULL with RANGE partitioning.  If you insert a row into a table partitioned by RANGE such that the column value used to determine the partition is NULL, the row is inserted into the lowest partition. For example, consider these two tables in a database named p, created as follows:

mysql> CREATE TABLE t1 (
    ->     c1 INT,
    ->     c2 VARCHAR(20)
    -> )
    -> PARTITION BY RANGE(c1) (
    ->     PARTITION p0 VALUES LESS THAN (0),
    ->     PARTITION p1 VALUES LESS THAN (10),
    ->     PARTITION p2 VALUES LESS THAN MAXVALUE
    -> );
Query OK, 0 rows affected (0.09 sec)

mysql> CREATE TABLE t2 (
    ->     c1 INT,
    ->     c2 VARCHAR(20)
    -> )
    -> PARTITION BY RANGE(c1) (
    ->     PARTITION p0 VALUES LESS THAN (-5),
    ->     PARTITION p1 VALUES LESS THAN (0),
    ->     PARTITION p2 VALUES LESS THAN (10),
    ->     PARTITION p3 VALUES LESS THAN MAXVALUE
    -> );
Query OK, 0 rows affected (0.09 sec)

You can see the partitions created by these two CREATE TABLE statements using the following query against the PARTITIONS table in the INFORMATION_SCHEMA database:

mysql> SELECT TABLE_NAME, PARTITION_NAME, TABLE_ROWS, AVG_ROW_LENGTH, DATA_LENGTH
     >   FROM INFORMATION_SCHEMA.PARTITIONS
     >   WHERE TABLE_SCHEMA = 'p' AND TABLE_NAME LIKE 't_';
+------------+----------------+------------+----------------+-------------+
| TABLE_NAME | PARTITION_NAME | TABLE_ROWS | AVG_ROW_LENGTH | DATA_LENGTH |
+------------+----------------+------------+----------------+-------------+
| t1         | p0             |          0 |              0 |           0 |
| t1         | p1             |          0 |              0 |           0 |
| t1         | p2             |          0 |              0 |           0 |
| t2         | p0             |          0 |              0 |           0 |
| t2         | p1             |          0 |              0 |           0 |
| t2         | p2             |          0 |              0 |           0 |
| t2         | p3             |          0 |              0 |           0 |
+------------+----------------+------------+----------------+-------------+
7 rows in set (0.00 sec)

(For more information about this table, see Section 19.19, “The INFORMATION_SCHEMA PARTITIONS Table”.) Now let us populate each of these tables with a single row containing a NULL in the column used as the partitioning key, and verify that the rows were inserted using a pair of SELECT statements:

mysql> INSERT INTO t1 VALUES (NULL, 'mothra');
Query OK, 1 row affected (0.00 sec)

mysql> INSERT INTO t2 VALUES (NULL, 'mothra');
Query OK, 1 row affected (0.00 sec)

mysql> SELECT * FROM t1;
+------+--------+
| id   | name   |
+------+--------+
| NULL | mothra |
+------+--------+
1 row in set (0.00 sec)

mysql> SELECT * FROM t2;
+------+--------+
| id   | name   |
+------+--------+
| NULL | mothra |
+------+--------+
1 row in set (0.00 sec)

You can see which partitions are used to store the inserted rows by rerunning the previous query against INFORMATION_SCHEMA.PARTITIONS and inspecting the output:

mysql> SELECT TABLE_NAME, PARTITION_NAME, TABLE_ROWS, AVG_ROW_LENGTH, DATA_LENGTH
     >   FROM INFORMATION_SCHEMA.PARTITIONS
     >   WHERE TABLE_SCHEMA = 'p' AND TABLE_NAME LIKE 't_';
+------------+----------------+------------+----------------+-------------+
| TABLE_NAME | PARTITION_NAME | TABLE_ROWS | AVG_ROW_LENGTH | DATA_LENGTH |
+------------+----------------+------------+----------------+-------------+
| t1         | p0             |          1 |             20 |          20 |
| t1         | p1             |          0 |              0 |           0 |
| t1         | p2             |          0 |              0 |           0 |
| t2         | p0             |          1 |             20 |          20 |
| t2         | p1             |          0 |              0 |           0 |
| t2         | p2             |          0 |              0 |           0 |
| t2         | p3             |          0 |              0 |           0 |
+------------+----------------+------------+----------------+-------------+
7 rows in set (0.01 sec)

You can also demonstrate that these rows were stored in the lowest partition of each table by dropping these partitions, and then re-running the SELECT statements:

mysql> ALTER TABLE t1 DROP PARTITION p0;
Query OK, 0 rows affected (0.16 sec)

mysql> ALTER TABLE t2 DROP PARTITION p0;
Query OK, 0 rows affected (0.16 sec)

mysql> SELECT * FROM t1;
Empty set (0.00 sec)

mysql> SELECT * FROM t2;
Empty set (0.00 sec)

(For more information on ALTER TABLE ... DROP PARTITION, see Section 12.1.6, “ALTER TABLE Syntax”.)

NULL is also treated in this way for partitioning expressions that use SQL functions. Suppose that we define a table using a CREATE TABLE statement such as this one:

CREATE TABLE tndate (
    id INT,
    dt DATE
)
PARTITION BY RANGE( YEAR(dt) ) (
    PARTITION p0 VALUES LESS THAN (1990),
    PARTITION p1 VALUES LESS THAN (2000),
    PARTITION p2 VALUES LESS THAN MAXVALUE
);

As with other MySQL functions, YEAR(NULL) returns NULL. A row with a dt column value of NULL is treated as though the partitioning expression evaluated to a value less than any other value, and so is inserted into partition p0.

Handling of NULL with LIST partitioning.  A table that is partitioned by LIST admits NULL values if and only if one of its partitions is defined using that value-list that contains NULL. The converse of this is that a table partitioned by LIST which does not explicitly use NULL in a value list rejects rows resulting in a NULL value for the partitioning expression, as shown in this example:

mysql> CREATE TABLE ts1 (
    ->     c1 INT,
    ->     c2 VARCHAR(20)
    -> )
    -> PARTITION BY LIST(c1) (
    ->     PARTITION p0 VALUES IN (0, 3, 6),
    ->     PARTITION p1 VALUES IN (1, 4, 7),
    ->     PARTITION p2 VALUES IN (2, 5, 8)
    -> );
Query OK, 0 rows affected (0.01 sec)

mysql> INSERT INTO ts1 VALUES (9, 'mothra');
ERROR 1504 (HY000): Table has no partition for value 9

mysql> INSERT INTO ts1 VALUES (NULL, 'mothra');
ERROR 1504 (HY000): Table has no partition for value NULL

Only rows having a c1 value between 0 and 8 inclusive can be inserted into ts1. NULL falls outside this range, just like the number 9. We can create tables ts2 and ts3 having value lists containing NULL, as shown here:

mysql> CREATE TABLE ts2 (
    ->     c1 INT,
    ->     c2 VARCHAR(20)
    -> )
    -> PARTITION BY LIST(c1) (
    ->     PARTITION p0 VALUES IN (0, 3, 6),
    ->     PARTITION p1 VALUES IN (1, 4, 7),
    ->     PARTITION p2 VALUES IN (2, 5, 8),
    ->     PARTITION p3 VALUES IN (NULL)
    -> );
Query OK, 0 rows affected (0.01 sec)

mysql> CREATE TABLE ts3 (
    ->     c1 INT,
    ->     c2 VARCHAR(20)
    -> )
    -> PARTITION BY LIST(c1) (
    ->     PARTITION p0 VALUES IN (0, 3, 6),
    ->     PARTITION p1 VALUES IN (1, 4, 7, NULL),
    ->     PARTITION p2 VALUES IN (2, 5, 8)
    -> );
Query OK, 0 rows affected (0.01 sec)

When defining value lists for partitioning, you can (and should) treat NULL just as you would any other value. For example, both VALUES IN (NULL) and VALUES IN (1, 4, 7, NULL) are valid, as are VALUES IN (1, NULL, 4, 7), VALUES IN (NULL, 1, 4, 7), and so on. You can insert a row having NULL for column c1 into each of the tables ts2 and ts3:

mysql> INSERT INTO ts2 VALUES (NULL, 'mothra');
Query OK, 1 row affected (0.00 sec)

mysql> INSERT INTO ts3 VALUES (NULL, 'mothra');
Query OK, 1 row affected (0.00 sec)

By issuing the appropriate query against INFORMATION_SCHEMA.PARTITIONS, you can determine which partitions were used to store the rows just inserted (we assume, as in the previous examples, that the partitioned tables were created in the p database):

mysql> SELECT TABLE_NAME, PARTITION_NAME, TABLE_ROWS, AVG_ROW_LENGTH, DATA_LENGTH
     >   FROM INFORMATION_SCHEMA.PARTITIONS
     >   WHERE TABLE_SCHEMA = 'p' AND TABLE_NAME LIKE 'ts_';
+------------+----------------+------------+----------------+-------------+
| TABLE_NAME | PARTITION_NAME | TABLE_ROWS | AVG_ROW_LENGTH | DATA_LENGTH |
+------------+----------------+------------+----------------+-------------+
| ts2        | p0             |          0 |              0 |           0 |
| ts2        | p1             |          0 |              0 |           0 |
| ts2        | p2             |          0 |              0 |           0 |
| ts2        | p3             |          1 |             20 |          20 |
| ts3        | p0             |          0 |              0 |           0 |
| ts3        | p1             |          1 |             20 |          20 |
| ts3        | p2             |          0 |              0 |           0 |
+------------+----------------+------------+----------------+-------------+
7 rows in set (0.01 sec)

As shown earlier in this section, you can also verify which partitions were used for storing the rows by deleting these partitions and then performing a SELECT.

Handling of NULL with HASH and KEY partitioning.  NULL is handled somewhat differently for tables partitioned by HASH or KEY. In these cases, any partition expression that yields a NULL value is treated as though its return value were zero. We can verify this behavior by examining the effects on the file system of creating a table partitioned by HASH and populating it with a record containing appropriate values. Suppose that you have a table th (also in the p database) created using the following statement:

mysql> CREATE TABLE th (
    ->     c1 INT,
    ->     c2 VARCHAR(20)
    -> )
    -> PARTITION BY HASH(c1)
    -> PARTITIONS 2;
Query OK, 0 rows affected (0.00 sec)

The partitions belonging to this table can be viewed like this:

mysql> SELECT TABLE_NAME,PARTITION_NAME,TABLE_ROWS,AVG_ROW_LENGTH,DATA_LENGTH
     >   FROM INFORMATION_SCHEMA.PARTITIONS
     >   WHERE TABLE_SCHEMA = 'p' AND TABLE_NAME ='th';
+------------+----------------+------------+----------------+-------------+
| TABLE_NAME | PARTITION_NAME | TABLE_ROWS | AVG_ROW_LENGTH | DATA_LENGTH |
+------------+----------------+------------+----------------+-------------+
| th         | p0             |          0 |              0 |           0 |
| th         | p1             |          0 |              0 |           0 |
+------------+----------------+------------+----------------+-------------+
2 rows in set (0.00 sec)

Note that TABLE_ROWS for each partition is 0. Now insert two rows into th whose c1 column values are NULL and 0, and verify that these rows were inserted:

mysql> INSERT INTO th VALUES (NULL, 'mothra'), (0, 'gigan');
Query OK, 1 row affected (0.00 sec)

mysql> SELECT * FROM th;
+------+---------+
| c1   | c2      |
+------+---------+
| NULL | mothra  |
+------+---------+
|    0 | gigan   |
+------+---------+
2 rows in set (0.01 sec)

Recall that for any integer N, the value of NULL MOD N is always NULL. For tables that are partitioned by HASH or KEY, this result is treated for determining the correct partition as 0. Checking the INFORMATION_SCHEMA.PARTITIONS table once again, we can see that both rows were inserted into partition p0:

mysql> SELECT TABLE_NAME, PARTITION_NAME, TABLE_ROWS, AVG_ROW_LENGTH, DATA_LENGTH
     >   FROM INFORMATION_SCHEMA.PARTITIONS
     >   WHERE TABLE_SCHEMA = 'p' AND TABLE_NAME ='th';
+------------+----------------+------------+----------------+-------------+
| TABLE_NAME | PARTITION_NAME | TABLE_ROWS | AVG_ROW_LENGTH | DATA_LENGTH |
+------------+----------------+------------+----------------+-------------+
| th         | p0             |          2 |             20 |          20 |
| th         | p1             |          0 |              0 |           0 |
+------------+----------------+------------+----------------+-------------+
2 rows in set (0.00 sec)

If you repeat this example using PARTITION BY KEY in place of PARTITION BY HASH in the definition of the table, you can verify easily that NULL is also treated like 0 for this type of partitioning as well.

Range and list partitions are very similar with regard to how the adding and dropping of partitions are handled. For this reason we discuss the management of both sorts of partitioning in this section. For information about working with tables that are partitioned by hash or key, see Section 17.3.2, “Management of HASH and KEY Partitions”. Dropping a RANGE or LIST partition is more straightforward than adding one, so we discuss this first.

Dropping a partition from a table that is partitioned by either RANGE or by LIST can be accomplished using the ALTER TABLE statement with a DROP PARTITION clause. Here is a very basic example, which supposes that you have already created a table which is partitioned by range and then populated with 10 records using the following CREATE TABLE and INSERT statements:

mysql> CREATE TABLE tr (id INT, name VARCHAR(50), purchased DATE)
    ->     PARTITION BY RANGE( YEAR(purchased) ) (
    ->         PARTITION p0 VALUES LESS THAN (1990),
    ->         PARTITION p1 VALUES LESS THAN (1995),
    ->         PARTITION p2 VALUES LESS THAN (2000),
    ->         PARTITION p3 VALUES LESS THAN (2005)
    ->     );
Query OK, 0 rows affected (0.01 sec)

mysql> INSERT INTO tr VALUES
    ->     (1, 'desk organiser', '2003-10-15'),
    ->     (2, 'CD player', '1993-11-05'),
    ->     (3, 'TV set', '1996-03-10'),
    ->     (4, 'bookcase', '1982-01-10'),
    ->     (5, 'exercise bike', '2004-05-09'),
    ->     (6, 'sofa', '1987-06-05'),
    ->     (7, 'popcorn maker', '2001-11-22'),
    ->     (8, 'aquarium', '1992-08-04'),
    ->     (9, 'study desk', '1984-09-16'),
    ->     (10, 'lava lamp', '1998-12-25');
Query OK, 10 rows affected (0.01 sec)

You can see which items should have been inserted into partition p2 as shown here:

mysql> SELECT * FROM tr
    -> WHERE purchased BETWEEN '1995-01-01' AND '1999-12-31';
+------+-----------+------------+
| id   | name      | purchased  |
+------+-----------+------------+
|    3 | TV set    | 1996-03-10 |
|   10 | lava lamp | 1998-12-25 |
+------+-----------+------------+
2 rows in set (0.00 sec)

To drop the partition named p2, execute the following command:

mysql> ALTER TABLE tr DROP PARTITION p2;
Query OK, 0 rows affected (0.03 sec)

It is very important to remember that, when you drop a partition, you also delete all the data that was stored in that partition. You can see that this is the case by re-running the previous SELECT query:

mysql> SELECT * FROM tr WHERE purchased
    -> BETWEEN '1995-01-01' AND '1999-12-31';
Empty set (0.00 sec)

Because of this, you must have the DROP privilege for a table before you can execute ALTER TABLE ... DROP PARTITION on that table.

If you wish to drop all data from all partitions while preserving the table definition and its partitioning scheme, use the TRUNCATE TABLE command. (See Section 12.2.11, “TRUNCATE TABLE Syntax”.)

If you intend to change the partitioning of a table without losing data, use ALTER TABLE ... REORGANIZE PARTITION instead. See below or in Section 12.1.6, “ALTER TABLE Syntax”, for information about REORGANIZE PARTITION.

If you now execute a SHOW CREATE TABLE command, you can see how the partitioning makeup of the table has been changed:

mysql> SHOW CREATE TABLE tr\G
*************************** 1. row ***************************
       Table: tr
Create Table: CREATE TABLE `tr` (
  `id` int(11) default NULL,
  `name` varchar(50) default NULL,
  `purchased` date default NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1
PARTITION BY RANGE ( YEAR(purchased) ) (
  PARTITION p0 VALUES LESS THAN (1990) ENGINE = MyISAM,
  PARTITION p1 VALUES LESS THAN (1995) ENGINE = MyISAM,
  PARTITION p3 VALUES LESS THAN (2005) ENGINE = MyISAM
)
1 row in set (0.01 sec)

When you insert new rows into the changed table with purchased column values between '1995-01-01' and '2004-12-31' inclusive, those rows will be stored in partition p3. You can verify this as follows:

mysql> INSERT INTO tr VALUES (11, 'pencil holder', '1995-07-12');
Query OK, 1 row affected (0.00 sec)

mysql> SELECT * FROM tr WHERE purchased
    -> BETWEEN '1995-01-01' AND '2004-12-31';
+------+----------------+------------+
| id   | name           | purchased  |
+------+----------------+------------+
|   11 | pencil holder  | 1995-07-12 |
|    1 | desk organiser | 2003-10-15 |
|    5 | exercise bike  | 2004-05-09 |
|    7 | popcorn maker  | 2001-11-22 |
+------+----------------+------------+
4 rows in set (0.00 sec)

mysql> ALTER TABLE tr DROP PARTITION p3;
Query OK, 0 rows affected (0.03 sec)

mysql> SELECT * FROM tr WHERE purchased
    -> BETWEEN '1995-01-01' AND '2004-12-31';
Empty set (0.00 sec)

Note that the number of rows dropped from the table as a result of ALTER TABLE ... DROP PARTITION is not reported by the server as it would be by the equivalent DELETE query.

Dropping LIST partitions uses exactly the same ALTER TABLE ... DROP PARTITION syntax as used for dropping RANGE partitions. However, there is one important difference in the effect this has on your use of the table afterward: You can no longer insert into the table any rows having any of the values that were included in the value list defining the deleted partition. (See Section 17.2.2, “LIST Partitioning”, for an example.)

To add a new range or list partition to a previously partitioned table, use the ALTER TABLE ... ADD PARTITION statement. For tables which are partitioned by RANGE, this can be used to add a new range to the end of the list of existing partitions. Suppose that you have a partitioned table containing membership data for your organisation, which is defined as follows:

CREATE TABLE members (
    id INT,
    fname VARCHAR(25),
    lname VARCHAR(25),
    dob DATE
)
PARTITION BY RANGE( YEAR(dob) ) (
    PARTITION p0 VALUES LESS THAN (1970),
    PARTITION p1 VALUES LESS THAN (1980),
    PARTITION p2 VALUES LESS THAN (1990)
);

Suppose further that the minimum age for members is 16. As the calendar approaches the end of 2005, you realize that you will soon be admitting members who were born in 1990 (and later in years to come). You can modify the members table to accommodate new members born in the years 1990–1999 as shown here:

ALTER TABLE ADD PARTITION (PARTITION p3 VALUES LESS THAN (2000));

mysql> ALTER TABLE members
     >     ADD PARTITION (
     >     PARTITION p3 VALUES LESS THAN (1960));
ERROR 1463 (HY000): VALUES LESS THAN value must be strictly »
   increasing for each partition

In a similar fashion, you can add new partitions to a table that is partitioned by LIST. For example, given a table defined like so:

CREATE TABLE tt (
    id INT,
    data INT
)
PARTITION BY LIST(data) (
    PARTITION p0 VALUES IN (5, 10, 15),
    PARTITION p1 VALUES IN (6, 12, 18)
);

You can add a new partition in which to store rows having the data column values 7, 14, and 21 as shown:

ALTER TABLE tt ADD PARTITION (PARTITION p2 VALUES IN (7, 14, 21));

Note that you cannot add a new LIST partition encompassing any values that are already included in the value list of an existing partition. If you attempt to do so, an error will result:

mysql> ALTER TABLE tt ADD PARTITION 
     >     (PARTITION np VALUES IN (4, 8, 12));
ERROR 1465 (HY000): Multiple definition of same constant »
                    in list partitioning

Because any rows with the data column value 12 have already been assigned to partition p1, you cannot create a new partition on table tt that includes 12 in its value list. To accomplish this, you could drop p1, and add np and then a new p1 with a modified definition. However, as discussed earlier, this would result in the loss of all data stored in p1 — and it is often the case that this is not what you really want to do. Another solution might appear to be to make a copy of the table with the new partitioning and to copy the data into it using CREATE TABLE ... SELECT ..., then drop the old table and rename the new one, but this could be very time-consuming when dealing with a large amounts of data. This also might not be feasible in situations where high availability is a requirement.

You can add multiple partitions in a single ALTER TABLE ... ADD PARTITION statement as shown here:

CREATE TABLE employees (
  id INT NOT NULL,
  fname VARCHAR(50) NOT NULL,
  lname VARCHAR(50) NOT NULL,
  hired DATE NOT NULL
)
PARTITION BY RANGE( YEAR(hired) ) (
  PARTITION p1 VALUES LESS THAN (1991),
  PARTITION p2 VALUES LESS THAN (1996),
  PARTITION p3 VALUES LESS THAN (2001),
  PARTITION p4 VALUES LESS THAN (2005)
);

ALTER TABLE employees ADD PARTITION (
    PARTITION p5 VALUES LESS THAN (2010),
    PARTITION p6 VALUES LESS THAN MAXVALUE
);

Fortunately, MySQL's partitioning implementation provides ways to redefine partitions without losing data. Let us look first at a couple of simple examples involving RANGE partitioning. Recall the members table which is now defined as shown here:

mysql> SHOW CREATE TABLE members\G
*************************** 1. row ***************************
       Table: members
Create Table: CREATE TABLE `members` (
  `id` int(11) default NULL,
  `fname` varchar(25) default NULL,
  `lname` varchar(25) default NULL,
  `dob` date default NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1
PARTITION BY RANGE ( YEAR(dob) ) (
  PARTITION p0 VALUES LESS THAN (1970) ENGINE = MyISAM,
  PARTITION p1 VALUES LESS THAN (1980) ENGINE = MyISAM,
  PARTITION p2 VALUES LESS THAN (1990) ENGINE = MyISAM.
  PARTITION p3 VALUES LESS THAN (2000) ENGINE = MyISAM
)

Suppose that you would like to move all rows representing members born before 1960 into a separate partition. As we have already seen, this cannot be done using ALTER TABLE ... ADD PARTITION. However, you can use another partition-related extension to ALTER TABLE in order to accomplish this:

ALTER TABLE members REORGANIZE PARTITION p0 INTO (
    PARTITION s0 VALUES LESS THAN (1960),
    PARTITION s1 VALUES LESS THAN (1970)
);

In effect, this command splits partition p0 into two new partitions s0 and s1. It also moves the data that was stored in p0 into the new partitions according to the rules embodied in the two PARTITION ... VALUES ... clauses, so that s0 contains only those records for which YEAR(dob) is less than 1960 and s1 contains those rows in which YEAR(dob) is greater than or equal to 1960 but less than 1970.

A REORGANIZE PARTITION clause may also be used for merging adjacent partitions. You can return the members table to its previous partitioning as shown here:

ALTER TABLE members REORGANIZE PARTITION s0,s1 INTO (
    PARTITION p0 VALUES LESS THAN (1970)
);

No data is lost in splitting or merging partitions using REORGANIZE PARTITION. In executing the above statement, MySQL moves all of the records that were stored in partitions s0 and s1 into partition p0.

The general syntax for REORGANIZE PARTITION is:

ALTER TABLE tbl_name
    REORGANIZE PARTITION partition_list
    INTO (partition_definitions);

Here, tbl_name is the name of the partitioned table, and partition_list is a comma-separated list of names of one or more existing partitions to be changed. partition_definitions is a comma-separated list of new partition definitions, which follow the same rules as for the partition_definitions list used in CREATE TABLE (see Section 12.1.14, “CREATE TABLE Syntax”). It should be noted that you are not limited to merging several partitions into one, or to splitting one partition into many, when using REORGANIZE PARTITION. For example, you can reorganize all four partitions of the members table into two, as follows:

ALTER TABLE members REORGANIZE PARTITION p0,p1,p2,p3 INTO (
    PARTITION m0 VALUES LESS THAN (1980),
    PARTITION m1 VALUES LESS THAN (2000)
);

You can also use REORGANIZE PARTITION with tables that are partitioned by LIST. Let us return to the problem of adding a new partition to the list-partitioned tt table and failing because the new partition had a value that was already present in the value-list of one of the existing partitions. We can handle this by adding a partition that contains only nonconflicting values, and then reorganizing the new partition and the existing one so that the value which was stored in the existing one is now moved to the new one:

ALTER TABLE tt ADD PARTITION (PARTITION np VALUES IN (4, 8));
ALTER TABLE tt REORGANIZE PARTITION p1,np INTO (
    PARTITION p1 VALUES IN (6, 18),
    PARTITION np VALUES in (4, 8, 12)
);

Here are some key points to keep in mind when using ALTER TABLE ... REORGANIZE PARTITION to repartition tables that are partitioned by RANGE or LIST:

Tables which are partitioned by hash or by key are very similar to one another with regard to making changes in a partitioning setup, and both differ in a number of ways from tables which have been partitioned by range or list. For that reason, this section addresses the modification of tables partitioned by hash or by key only. For a discussion of adding and dropping of partitions of tables that are partitioned by range or list, see Section 17.3.1, “Management of RANGE and LIST Partitions”.

You cannot drop partitions from tables that are partitioned by HASH or KEY in the same way that you can from tables that are partitioned by RANGE or LIST. However, you can merge HASH or KEY partitions using the ALTER TABLE ... COALESCE PARTITION command. Suppose that you have a table containing data about clients, which is divided into twelve partitions. The clients table is defined as shown here:

CREATE TABLE clients (
    id INT,
    fname VARCHAR(30),
    lname VARCHAR(30),
    signed DATE
)
PARTITION BY HASH( MONTH(signed) )
PARTITIONS 12;

To reduce the number of partitions from twelve to eight, execute the following ALTER TABLE command:

mysql> ALTER TABLE clients COALESCE PARTITION 4;
Query OK, 0 rows affected (0.02 sec)

COALESCE works equally well with tables that are partitioned by HASH, KEY, LINEAR HASH, or LINEAR KEY. Here is an example similar to the previous one, differing only in that the table is partitioned by LINEAR KEY:

mysql> CREATE TABLE clients_lk (
    ->     id INT,
    ->     fname VARCHAR(30),
    ->     lname VARCHAR(30),
    ->     signed DATE
    -> )
    -> PARTITION BY LINEAR KEY(signed)
    -> PARTITIONS 12;
Query OK, 0 rows affected (0.03 sec)

mysql> ALTER TABLE clients_lk COALESCE PARTITION 4;
Query OK, 0 rows affected (0.06 sec)
Records: 0  Duplicates: 0  Warnings: 0

Note that the number following COALESCE PARTITION is the number of partitions to merge into the remainder — in other words, it is the number of partitions to remove from the table.

If you attempt to remove more partitions than the table has, the result is an error like the one shown:

mysql> ALTER TABLE clients COALESCE PARTITION 18;
ERROR 1478 (HY000): Cannot remove all partitions, use DROP TABLE instead

To increase the number of partitions for the clients table from 12 to 18. use ALTER TABLE ... ADD PARTITION as shown here:

ALTER TABLE clients ADD PARTITION PARTITIONS 6;

A number of table and partition maintenance tasks can be carried out using SQL statements intended for such purposes on partitioned tables in MySQL 5.5.

Table maintenance of partitioned tables can be accomplished using the statements CHECK TABLE, OPTIMIZE TABLE, ANALYZE TABLE, and REPAIR TABLE, which are supported for partitioned tables.

You can use a number of extensions to ALTER TABLE for performing operations of this type on one or more partitions directly, as described in the following list:

Each of the statements in the list just shown also supports the keyword ALL in place of the list of partition names. Using ALL causes the statement to act on all partitions in the table.

Beginning with MySQL 5.5.0, you can also truncate partitions using ALTER TABLE ... TRUNCATE PARTITION. This statement can be used to delete all rows from one or more partitions in much the same way that TRUNCATE TABLE deletes all rows from a table.

ALTER TABLE ... TRUNCATE PARTITION ALL truncates all partitions in the table.

This section discusses obtaining information about existing partitions, which can be done in a number of ways. These include:

As discussed elsewhere in this chapter, SHOW CREATE TABLE includes in its output the PARTITION BY clause used to create a partitioned table. For example:

mysql> SHOW CREATE TABLE trb3\G
*************************** 1. row ***************************
       Table: trb3
Create Table: CREATE TABLE `trb3` (
  `id` int(11) default NULL,
  `name` varchar(50) default NULL,
  `purchased` date default NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1
PARTITION BY RANGE (YEAR(purchased)) (
  PARTITION p0 VALUES LESS THAN (1990) ENGINE = MyISAM,
  PARTITION p1 VALUES LESS THAN (1995) ENGINE = MyISAM,
  PARTITION p2 VALUES LESS THAN (2000) ENGINE = MyISAM,
  PARTITION p3 VALUES LESS THAN (2005) ENGINE = MyISAM
)
1 row in set (0.00 sec)

The output from SHOW TABLE STATUS for partitioned tables is the same as that for nonpartitioned tables, except that the Create_options column contains the string partitioned. The Engine column contains the name of the storage engine used by all partitions of the table. (See Section 12.5.5.37, “SHOW TABLE STATUS Syntax”, for more information about this statement.)

You can also obtain information about partitions from INFORMATION_SCHEMA, which contains a PARTITIONS table. See Section 19.19, “The INFORMATION_SCHEMA PARTITIONS Table”.

It is possible to determine which partitions of a partitioned table are involved in a given SELECT query using EXPLAIN PARTITIONS. The PARTITIONS keyword adds a partitions column to the output of EXPLAIN listing the partitions from which records would be matched by the query.

Suppose that you have a table trb1 defined and populated as follows:

CREATE TABLE trb1 (id INT, name VARCHAR(50), purchased DATE)
    PARTITION BY RANGE(id)
    (
        PARTITION p0 VALUES LESS THAN (3),
        PARTITION p1 VALUES LESS THAN (7),
        PARTITION p2 VALUES LESS THAN (9),
        PARTITION p3 VALUES LESS THAN (11)
    );

INSERT INTO trb1 VALUES
    (1, 'desk organiser', '2003-10-15'),
    (2, 'CD player', '1993-11-05'),
    (3, 'TV set', '1996-03-10'),
    (4, 'bookcase', '1982-01-10'),
    (5, 'exercise bike', '2004-05-09'),
    (6, 'sofa', '1987-06-05'),
    (7, 'popcorn maker', '2001-11-22'),
    (8, 'aquarium', '1992-08-04'),
    (9, 'study desk', '1984-09-16'),
    (10, 'lava lamp', '1998-12-25');

You can see which partitions are used in a query such as SELECT * FROM trb1;, as shown here:

mysql> EXPLAIN PARTITIONS SELECT * FROM trb1\G
*************************** 1. row ***************************
           id: 1
  select_type: SIMPLE
        table: trb1
   partitions: p0,p1,p2,p3
         type: ALL
possible_keys: NULL
          key: NULL
      key_len: NULL
          ref: NULL
         rows: 10
        Extra: Using filesort

In this case, all four partitions are searched. However, when a limiting condition making use of the partitioning key is added to the query, you can see that only those partitions containing matching values are scanned, as shown here:

mysql> EXPLAIN PARTITIONS SELECT * FROM trb1 WHERE id < 5\G
*************************** 1. row ***************************
           id: 1
  select_type: SIMPLE
        table: trb1
   partitions: p0,p1
         type: ALL
possible_keys: NULL
          key: NULL
      key_len: NULL
          ref: NULL
         rows: 10
        Extra: Using where

EXPLAIN PARTITIONS provides information about keys used and possible keys, just as with the standard EXPLAIN SELECT statement:

mysql> ALTER TABLE trb1 ADD PRIMARY KEY (id);
Query OK, 10 rows affected (0.03 sec)
Records: 10  Duplicates: 0  Warnings: 0

mysql> EXPLAIN PARTITIONS SELECT * FROM trb1 WHERE id < 5\G
*************************** 1. row ***************************
           id: 1
  select_type: SIMPLE
        table: trb1
   partitions: p0,p1
         type: range
possible_keys: PRIMARY
          key: PRIMARY
      key_len: 4
          ref: NULL
         rows: 7
        Extra: Using where

You should take note of the following restrictions and limitations on EXPLAIN PARTITIONS:

The rows column of EXPLAIN PARTITIONS output displays the total number of records in the table.

See also Section 12.3.2, “EXPLAIN Syntax”.

This section discusses the relationship of partitioning keys with primary keys and unique keys. The rule governing this relationship can be expressed as follows: All columns used in the partitioning expression for a partitioned table must be part of every unique key that the table may have.

In other words, every unique key on the table must use every column in the table's partitioning expression. (This also includes the table's primary key, since it is by definition a unique key. This particular case is discussed later in this section.) For example, each of the following table creation statements is invalid:

CREATE TABLE t1 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    UNIQUE KEY (col1, col2)
)
PARTITION BY HASH(col3)
PARTITIONS 4;

CREATE TABLE t2 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    UNIQUE KEY (col1),
    UNIQUE KEY (col3)
)
PARTITION BY HASH(col1 + col3)
PARTITIONS 4;

In each case, the proposed table would have at least one unique key that does not include all columns used in the partitioning expression.

Each of the following statements is valid, and represents one way in which the corresponding invalid table creation statement could be made to work:

CREATE TABLE t1 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    UNIQUE KEY (col1, col2, col3)
)
PARTITION BY HASH(col3)
PARTITIONS 4;

CREATE TABLE t2 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    UNIQUE KEY (col1, col3)
)
PARTITION BY HASH(col1 + col3)
PARTITIONS 4;

This example shows the error produced in such cases:

mysql> CREATE TABLE t3 (
    ->     col1 INT NOT NULL,
    ->     col2 DATE NOT NULL,
    ->     col3 INT NOT NULL,
    ->     col4 INT NOT NULL,
    ->     UNIQUE KEY (col1, col2),
    ->     UNIQUE KEY (col3)
    -> )
    -> PARTITION BY HASH(col1 + col3)
    -> PARTITIONS 4;
ERROR 1491 (HY000): A PRIMARY KEY must include all columns in the table's partitioning function

The CREATE TABLE statement fails because both col1 and col3 are included in the proposed partitioning key, but neither of these columns is part of both of unique keys on the table. This shows one possible fix for the invalid table definition:

mysql> CREATE TABLE t3 (
    ->     col1 INT NOT NULL,
    ->     col2 DATE NOT NULL,
    ->     col3 INT NOT NULL,
    ->     col4 INT NOT NULL,
    ->     UNIQUE KEY (col1, col2, col3),
    ->     UNIQUE KEY (col3)
    -> )
    -> PARTITION BY HASH(col3)
    -> PARTITIONS 4;
Query OK, 0 rows affected (0.05 sec)

In this case, the proposed partitioning key col3 is part of both unique keys, and the table creation statement succeeds.

The following table cannot be partitioned at all, because there is no way to include in a partitioning key any columns that belong to both unique keys:

CREATE TABLE t4 (
    col1 INT NOT NULL,
    col2 INT NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    UNQIUE KEY (col1, col3),
    UNQIUE KEY (col2, col4)
);

Since every primary key is by definition a unique key, this restriction also includes the table's primary key, if it has one. For example, the next two statements are invalid:

CREATE TABLE t5 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    PRIMARY KEY(col1, col2)
)
PARTITION BY HASH(col3)
PARTITIONS 4;

CREATE TABLE t6 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    PRIMARY KEY(col1, col3),
    UNIQUE KEY(col2)
)
PARTITION BY HASH( YEAR(col2) )
PARTITIONS 4;

In both cases, the primary key does not include all columns referenced in the partitioning expression. However, both of the next two statements are valid:

CREATE TABLE t7 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    PRIMARY KEY(col1, col2)
)
PARTITION BY HASH(col1 + YEAR(col2))
PARTITIONS 4;

CREATE TABLE t8 (
    col1 INT NOT NULL,
    col2 DATE NOT NULL,
    col3 INT NOT NULL,
    col4 INT NOT NULL,
    PRIMARY KEY(col1, col2, col4),
    UNIQUE KEY(col2, col1)
)
PARTITION BY HASH(col1 + YEAR(col2))
PARTITIONS 4;

If a table has no unique keys — this includes having no primary key — then this restriction does not apply, and you may use any column or columns in the partitioning expression as long as the column type is compatible with the partitioning type.

For the same reason, you cannot later add a unique key to a partitioned table unless the key includes all columns used by the table's partitioning expression. Consider given the partitioned table defined as shown here:

mysql> CREATE TABLE t_no_pk (c1 INT, c2 INT)
    ->     PARTITION BY RANGE(c1) (
    ->         PARTITION p0 VALUES LESS THAN (10),
    ->         PARTITION p1 VALUES LESS THAN (20),
    ->         PARTITION p2 VALUES LESS THAN (30),
    ->         PARTITION p3 VALUES LESS THAN (40)
    ->     );
Query OK, 0 rows affected (0.12 sec)

It is possible to add a primary key to t_no_pk using either of these ALTER TABLE statements:

#  possible PK
mysql> ALTER TABLE t_no_pk ADD PRIMARY KEY(c1);
Query OK, 0 rows affected (0.13 sec)
Records: 0  Duplicates: 0  Warnings: 0

# drop this PK
mysql> ALTER TABLE t_no_pk DROP PRIMARY KEY;
Query OK, 0 rows affected (0.10 sec)
Records: 0  Duplicates: 0  Warnings: 0

#  use another possible PK
mysql> ALTER TABLE t_no_pk ADD PRIMARY KEY(c1, c2);
Query OK, 0 rows affected (0.12 sec)
Records: 0  Duplicates: 0  Warnings: 0

# drop this PK
mysql> ALTER TABLE t_no_pk DROP PRIMARY KEY;
Query OK, 0 rows affected (0.09 sec)
Records: 0  Duplicates: 0  Warnings: 0

However, the next statement fails, because c1 is part of the partitioning key, but is not part of the proposed primary key:

#  fails with error 1503
mysql> ALTER TABLE t_no_pk ADD PRIMARY KEY(c2);
ERROR 1503 (HY000): A PRIMARY KEY must include all columns in the table's partitioning function

Since t_no_pk has only c1 in its partitioning expression, attempting to adding a unique key on c2 alone fails. However, you can add a unique key that uses both c1 and c2.

These rules also apply to existing nonpartitioned tables that you wish to partition using ALTER TABLE ... PARTITION BY. Consider a table np_pk defined as shown here:

mysql> CREATE TABLE np_pk (
    ->     id INT NOT NULL AUTO_INCREMENT,
    ->     name VARCHAR(50),
    ->     added DATE,
    ->     PRIMARY KEY (id)
    -> );
Query OK, 0 rows affected (0.08 sec)

The following ALTER TABLE statements fails with an error, because the added column is not part of any unique key in the table:

mysql> ALTER TABLE np_pk
    ->     PARTITION BY HASH( TO_DAYS(added) )
    ->     PARTITIONS 4;
ERROR 1503 (HY000): A PRIMARY KEY must include all columns in the table's partitioning function

However, this statement using the id column for the partitioning column is valid, as shown here:

mysql> ALTER TABLE np_pk
    ->     PARTITION BY HASH(id)
    ->     PARTITIONS 4;
Query OK, 0 rows affected (0.11 sec)
Records: 0  Duplicates: 0  Warnings: 0

In the case of np_pk, the only column that may be used as part of a partitioning expression is id; if you wish to partition this table using any other column or columns in the partitioning expression, you must first modify the table, either by adding the desired column or columns to the primary key, or by dropping the primary key altogether.

We are working to remove this limitation in a future MySQL release series.

The following limitations apply to the use of storage engines with user-defined partitioning of tables.

MERGE storage engine.  User-defined partitioning and the MERGE storage engine are not compatible. Tables using the MERGE storage engine cannot be partitioned. Partitioned tables cannot be merged.

FEDERATED storage engine.  Partitioning of FEDERATED tables is not supported; it is not possible to create partitioned FEDERATED tables. We are working to remove this limitation in a future MySQL release.

CSV storage engine.  Partitioned tables using the CSV storage engine are not supported; it is not possible to create partitioned CSV tables.

NDBCLUSTER storage engine (MySQL Cluster).  Partitioning by KEY (or LINEAR KEY) is the only type of partitioning supported for the NDBCLUSTER storage engine. It is not possible to create a MySQL Cluster table using any partitioning type other than [LINEAR] KEY, and attempting to do so fails with an error.

In addition, the maximum number of partitions that can be defined for an NDBCLUSTER table is 8 times the number of node groups in the cluster. (See MySQL Cluster Nodes, Node Groups, Replicas, and Partitions, for more information about node groups in MySQL Cluster.)

Upgrading partitioned tables.  When performing an upgrade, tables which are partitioned by KEY and which use any storage engine other than NDBCLUSTER must be dumped and reloaded.

Same storage engine for all partitions.  All partitions of a partitioned table must use the same storage engine and it must be the same storage engine used by the table as a whole. In addition, if one does not specify an engine on the table level, then one must do either of the following when creating or altering a partitioned table:

We are working to remove this limitation in a future MySQL release.

This section discusses limitations in MySQL Partitioning relating specifically to functions used in partitioning expressions.

Only the MySQL functions shown in the following table are supported in partitioning expressions:

CEILING() and FLOOR().  Each of these functions returns an integer only if it is passed an integer argument. This means, for example, that the following CREATE TABLE statement fails with an error, as shown here:

mysql> CREATE TABLE t (c FLOAT) PARTITION BY LIST( FLOOR(c) )(
    ->     PARTITION p0 VALUES IN (1,3,5),
    ->     PARTITION p1 VALUES IN (2,4,6)
    -> );
ERROR 1490 (HY000): The PARTITION function returns the wrong type

See Section 11.5.2, “Mathematical Functions”, for more information about the return types of these functions.