Partitioning
This document introduces TiDB's implementation of partitioning.
Partitioning types
This section introduces the types of partitioning which are available in TiDB. Currently, TiDB supports range partitioning and hash partitioning.
Range partitioning is used to resolve the performance issues caused by a large amount of deletions in the application, and it supports fast drop partition operations. Hash partitioning is used to scatter the data when there are a large amount of writes.
Range partitioning
When a table is partitioned by range, each partition contains rows for which the partitioning expression value lies within a given range. Ranges have to be contiguous but not overlapping. You can define it by using the VALUES LESS THAN
operation.
Assume you need to create a table that contains personnel records as follows:
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
);
You can partition a table by range in various ways as needed. For example, you can partition it by using the store_id
column:
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 partition scheme, all rows corresponding to employees whose store_id
is 1 through 5 are stored in the p0
partition while all employees whose store_id
is 6 through 10 are stored in p1
. Range partitioning requires the partitions to be ordered, from lowest to highest.
If you insert a row of data (72, 'Mitchell', 'Wilson', '1998-06-25', NULL, 13)
, it falls in the p2
partition. But if you insert a record whose store_id
is larger than 20, an error is reported because TiDB can not know which partition this record should be inserted into. In this case, you can use MAXVALUE
when creating a table:
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 larger than all other integer values. Now, all records whose store_id
is equal to or larger than 16 (the highest value defined) are stored in the p3
partition.
You can also partition a table by employees' job codes, which are the values of the job_code
column. Assume that two-digit job codes stand for regular employees, three-digit codes stand for office and customer support personnel, and four-digit codes stand for managerial personnel. Then you can create a partitioned table like this:
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 example, all rows relating to regular employees are stored in the p0
partition, all office and customer support personnel in the p1
partition, and all managerial personnel in the p2
partition.
Besides splitting up the table by store_id
, you can also partition a table by dates. For example, you can partition by employees' separation year:
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 range partitioning, you can partition based on the values of the timestamp
column and use the unix_timestamp()
function, for example:
CREATE TABLE quarterly_report_status (
report_id INT NOT NULL,
report_status VARCHAR(20) NOT NULL,
report_updated TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP
)
PARTITION BY RANGE ( UNIX_TIMESTAMP(report_updated) ) (
PARTITION p0 VALUES LESS THAN ( UNIX_TIMESTAMP('2008-01-01 00:00:00') ),
PARTITION p1 VALUES LESS THAN ( UNIX_TIMESTAMP('2008-04-01 00:00:00') ),
PARTITION p2 VALUES LESS THAN ( UNIX_TIMESTAMP('2008-07-01 00:00:00') ),
PARTITION p3 VALUES LESS THAN ( UNIX_TIMESTAMP('2008-10-01 00:00:00') ),
PARTITION p4 VALUES LESS THAN ( UNIX_TIMESTAMP('2009-01-01 00:00:00') ),
PARTITION p5 VALUES LESS THAN ( UNIX_TIMESTAMP('2009-04-01 00:00:00') ),
PARTITION p6 VALUES LESS THAN ( UNIX_TIMESTAMP('2009-07-01 00:00:00') ),
PARTITION p7 VALUES LESS THAN ( UNIX_TIMESTAMP('2009-10-01 00:00:00') ),
PARTITION p8 VALUES LESS THAN ( UNIX_TIMESTAMP('2010-01-01 00:00:00') ),
PARTITION p9 VALUES LESS THAN (MAXVALUE)
);
It is not allowed to use any other partitioning expression that contains the timestamp values.
Range partitioning is particularly useful when one or more of the following conditions are satisfied:
- You want to delete the old data. If you use the
employees
table in the previous example, you can delete all records of employees who left this company before the year 1991 by simply usingALTER TABLE employees DROP PARTITION p0;
. It is faster than executing theDELETE FROM employees WHERE YEAR(separated) <= 1990;
operation. - You want to use a column that contains time or date values, or containing values arising from some other series.
- You need to frequently run queries on the columns used for partitioning. For example, when executing a query like
EXPLAIN SELECT COUNT(*) FROM employees WHERE separated BETWEEN '2000-01-01' AND '2000-12-31' GROUP BY store_id;
, TiDB can quickly know that only the data in thep2
partition needs to be scanned, because the other partitions do not match theWHERE
condition.
Hash partitioning
Hash partitioning is used to make sure that data is evenly scattered into a certain number of partitions. With range partitioning, you must specify the range of the column values for each partition when you use range partitioning, while you just need to specify the number of partitions when you use hash partitioning.
Partitioning by hash requires you to append a PARTITION BY HASH (expr)
clause to the CREATE TABLE
statement. expr
is an expression that returns an integer. It can be a column name if the type of this column is integer. In addition, you might also need to append PARTITIONS num
, where num
is a positive integer indicating how many partitions a table is divided into.
The following operation creates a hash partitioned table, which is divided into 4 partitions by store_id
:
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 PARTITIONS num
is not specified, the default number of partitions is 1.
You can also use an SQL expression that returns an integer for expr
. For example, you can partition a table by the hire year:
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;
The most efficient hash 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.
For example, date_col
is a column whose type is DATE
, and the value of the TO_DAYS(date_col)
expression varies with the value of date_col
. YEAR(date_col)
is different from 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 still a good hash function, because its value varies in proportion to the value of date_col
.
In contrast, assume that you have an int_col
column whose type is INT
. Now consider about the expression POW(5-int_col,3) + 6
. It is not a good hash function though, because as the value of int_col
changes, the result of the expression does not change proportionally. A value change in int_col
might result in a huge change in the expression result. For example, when int_col
changes from 5 to 6, the change of the expression result is -1. But the result change might be -7 when int_col
changes from 6 to 7.
In conclusion, when the expression has a form that is closer to y = cx
, it is more suitable to be a hash function. Because the more non-linear an expression is, the more unevenly scattered the data among the partitions tends to be.
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 using PARTITIION BY HASH
, TiDB decides which partition the data should fall into based on the modulus of the result of the expression. In other words, if a partitioning expression is expr
and the number of partitions is num
, MOD(expr, num)
decides the partition in which the data is stored. Assume that t1
is defined as follows:
CREATE TABLE t1 (col1 INT, col2 CHAR(5), col3 DATE)
PARTITION BY HASH( YEAR(col3) )
PARTITIONS 4;
When you insert a row of data into t1
and the value of col3
is '2005-09-15', then this row is inserted into partition 1:
MOD(YEAR('2005-09-01'),4)
= MOD(2005,4)
= 1
How TiDB partitioning handles NULL
It is allowed in TiDB to use NULL
as the calculation result of a partitioning expression.
Handling of NULL with range partitioning
When you insert a row into a table partitioned by range, and the column value used to determine the partition is NULL
, then this row is inserted into the lowest partition.
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)
select * from t1 partition(p0);
+------|--------+
| c1 | c2 |
+------|--------+
| NULL | mothra |
+------|--------+
1 row in set (0.00 sec)
select * from t1 partition(p1);
Empty set (0.00 sec)
select * from t1 partition(p2);
Empty set (0.00 sec)
Drop the p0
partition and verify the result:
alter table t1 drop partition p0;
Query OK, 0 rows affected (0.08 sec)
select * from t1;
Empty set (0.00 sec)
Handling of NULL with hash partitioning
When partitioning tables by hash, there is a different way of handling NULL
value - if the calculation result of the partitioning expression is NULL
, it is considered as 0
.
CREATE TABLE th (
c1 INT,
c2 VARCHAR(20)
)
PARTITION BY HASH(c1)
PARTITIONS 2;
Query OK, 0 rows affected (0.00 sec)
INSERT INTO th VALUES (NULL, 'mothra'), (0, 'gigan');
Query OK, 2 rows affected (0.04 sec)
select * from th partition (p0);
+------|--------+
| c1 | c2 |
+------|--------+
| NULL | mothra |
| 0 | gigan |
+------|--------+
2 rows in set (0.00 sec)
select * from th partition (p1);
Empty set (0.00 sec)
You can see that the inserted record (NULL, 'mothra')
falls into the same partition as (0, 'gigan')
.
Partition management
You can add, drop, merge, split, redefine partitions by using ALTER TABLE
statements.
Range partition management
Create a partitioned table:
CREATE TABLE members (
id INT,
fname VARCHAR(25),
lname VARCHAR(25),
dob DATE
)
PARTITION BY RANGE( YEAR(dob) ) (
PARTITION p0 VALUES LESS THAN (1980),
PARTITION p1 VALUES LESS THAN (1990),
PARTITION p2 VALUES LESS THAN (2000)
);
Drop a partition:
ALTER TABLE members DROP PARTITION p2;
Query OK, 0 rows affected (0.03 sec)
Empty a partition:
ALTER TABLE members TRUNCATE PARTITION p1;
Query OK, 0 rows affected (0.03 sec)
Add a partition:
ALTER TABLE members ADD PARTITION (PARTITION p3 VALUES LESS THAN (2010));
When partitioning tables by range, ADD PARTITION
can be only appended to the very end of a partition list. If it is appended to an existing partition range, an error is reported:
ALTER TABLE members
ADD PARTITION (
PARTITION n VALUES LESS THAN (1970));
ERROR 1463 (HY000): VALUES LESS THAN value must be strictly »
increasing for each partition
Hash partition management
Unlike range partitioning, DROP PARTITION
is not supported in hash partitioning.
Currently, ALTER TABLE ... COALESCE PARTITION
is not supported in TiDB as well.
Partition pruning
Partition pruning is an optimization which is based on a very simple idea - do not scan the partitions that do not match.
Assume that you create a partitioned table t1
:
CREATE TABLE t1 (
fname VARCHAR(50) NOT NULL,
lname VARCHAR(50) NOT NULL,
region_code TINYINT UNSIGNED NOT NULL,
dob DATE NOT NULL
)
PARTITION BY RANGE( region_code ) (
PARTITION p0 VALUES LESS THAN (64),
PARTITION p1 VALUES LESS THAN (128),
PARTITION p2 VALUES LESS THAN (192),
PARTITION p3 VALUES LESS THAN MAXVALUE
);
If you want to get the result of this SELECT
statement:
SELECT fname, lname, region_code, dob
FROM t1
WHERE region_code > 125 AND region_code < 130;
It is evident that the result falls in either the p1
or the p2
partition, that is, you just need to search for the matching rows in p1
and p2
. Excluding the unneeded partitions is so-called "pruning". If the optimizer is able to prune a part of partitions, the execution of the query in the partitioned table will be much faster than that in a non-partitioned table.
The optimizer can prune partitions through WHERE
conditions in the following two scenarios:
- partition_column = constant
- partition_column IN (constant1, constant2, ..., constantN)
Some cases for partition pruning to take effect
Partition pruning uses the query conditions on the partitioned table, so if the query conditions can not be pushed down to the partitioned table according to the planner's optimization rules, partition pruning does not apply for this query.
For example:
create table t1 (x int) partition by range (x) ( partition p0 values less than (5), partition p1 values less than (10)); create table t2 (x int);explain select * from t1 left join t2 on t1.x = t2.x where t2.x > 5;In this query, the left out join is converted to the inner join, and then
t1.x > 5
is derived fromt1.x = t2.x
andt2.x > 5
, so it could be used in partition pruning and only the partitionp1
remains.explain select * from t1 left join t2 on t1.x = t2.x and t2.x > 5;In this query,
t2.x > 5
can not be pushed down to thet1
partitioned table, so partition pruning would not take effect for this query.Since partition pruning is done during the plan optimizing phase, it does not apply for those cases that filter conditions are unknown until the execution phase.
For example:
create table t1 (x int) partition by range (x) ( partition p0 values less than (5), partition p1 values less than (10));explain select * from t2 where x < (select * from t1 where t2.x < t1.x and t2.x < 2);This query reads a row from
t2
and uses the result for the subquery ont1
. Theoretically, partition pruning could benefit fromt1.x > val
expression in the subquery, but it does not take effect there as that happens in the execution phase.As a result of a limitation from current implementation, if a query condition can not be pushed down to TiKV, it can not be used by the partition pruning.
Take the
fn(col)
expression as an example. If the TiKV coprocessor supports thisfn
function,fn(col)
may be pushed down to the the leaf node (that is, partitioned table) according to the predicate push-down rule during the plan optimizing phase, and partition pruning can use it.If the TiKV coprocessor does not support this
fn
function,fn(col)
would not be pushed down to the leaf node. Instead, it becomes aSelection
node above the leaf node. The current partition pruning implementation does not support this kind of plan tree.For hash partition, the only query supported by partition pruning is the equal condition.
For range partition, for partition pruning to take effect, the partition expression must be in those forms:
col
orfn(col)
, and the query condition must be one of>
,<
,=
,>=
, and<=
. If the partition expression is in the form offn(col)
, thefn
function must be monotonous.If the
fn
function is monotonous, for anyx
andy
, ifx > y
, thenfn(x) > fn(y)
. Then thisfn
function can be called strictly monotonous. For anyx
andy
, ifx > y
, thenfn(x) >= fn(y)
. In this case,fn
could also be called "monotonous". In theory, all monotonous functions are supported by partition pruning.Currently, partition pruning in TiDB only support those monotonous functions:
unix_timestamp to_daysFor example, the partition expression is a simple column:
create table t (id int) partition by range (id) ( partition p0 values less than (5), partition p1 values less than (10)); select * from t where t > 6;Or the partition expression is in the form of
fn(col)
wherefn
isto_days
:create table t (dt datetime) partition by range (to_days(id)) ( partition p0 values less than (to_days('2020-04-01')), partition p1 values less than (to_days('2020-05-01'))); select * from t where t > '2020-04-18';An exception is
floor(unix_timestamp())
as the partition expression. TiDB does some optimization for that case by case, so it is supported by partition pruning.create table t (ts timestamp(3) not null default current_timestamp(3)) partition by range (floor(unix_timestamp(ts))) ( partition p0 values less than (unix_timestamp('2020-04-01 00:00:00')), partition p1 values less than (unix_timestamp('2020-05-01 00:00:00'))); select * from t where t > '2020-04-18 02:00:42.123';
Partition selection
SELECT
statements support partition selection, which is implemented by using a PARTITION
option.
CREATE TABLE employees (
id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
fname VARCHAR(25) NOT NULL,
lname VARCHAR(25) NOT NULL,
store_id INT NOT NULL,
department_id INT NOT NULL
)
PARTITION BY RANGE(id) (
PARTITION p0 VALUES LESS THAN (5),
PARTITION p1 VALUES LESS THAN (10),
PARTITION p2 VALUES LESS THAN (15),
PARTITION p3 VALUES LESS THAN MAXVALUE
);
INSERT INTO employees VALUES
('', 'Bob', 'Taylor', 3, 2), ('', 'Frank', 'Williams', 1, 2),
('', 'Ellen', 'Johnson', 3, 4), ('', 'Jim', 'Smith', 2, 4),
('', 'Mary', 'Jones', 1, 1), ('', 'Linda', 'Black', 2, 3),
('', 'Ed', 'Jones', 2, 1), ('', 'June', 'Wilson', 3, 1),
('', 'Andy', 'Smith', 1, 3), ('', 'Lou', 'Waters', 2, 4),
('', 'Jill', 'Stone', 1, 4), ('', 'Roger', 'White', 3, 2),
('', 'Howard', 'Andrews', 1, 2), ('', 'Fred', 'Goldberg', 3, 3),
('', 'Barbara', 'Brown', 2, 3), ('', 'Alice', 'Rogers', 2, 2),
('', 'Mark', 'Morgan', 3, 3), ('', 'Karen', 'Cole', 3, 2);
You can view the rows stored in the p1
partition:
SELECT * FROM employees PARTITION (p1);
+----|-------|--------|----------|---------------+
| id | fname | lname | store_id | department_id |
+----|-------|--------|----------|---------------+
| 5 | Mary | Jones | 1 | 1 |
| 6 | Linda | Black | 2 | 3 |
| 7 | Ed | Jones | 2 | 1 |
| 8 | June | Wilson | 3 | 1 |
| 9 | Andy | Smith | 1 | 3 |
+----|-------|--------|----------|---------------+
5 rows in set (0.00 sec)
If you want to get the rows in multiple partitions, you can use a list of partition names which are separated by commas. For example, SELECT * FROM employees PARTITION (p1, p2)
returns all rows in the p1
and p2
partitions.
When you use partition selection, you can still use WHERE
conditions and options such as ORDER BY
and LIMIT
. It is also supported to use aggregation options such as HAVING
and GROUP BY
.
SELECT * FROM employees PARTITION (p0, p2)
WHERE lname LIKE 'S%';
+----|-------|-------|----------|---------------+
| id | fname | lname | store_id | department_id |
+----|-------|-------|----------|---------------+
| 4 | Jim | Smith | 2 | 4 |
| 11 | Jill | Stone | 1 | 4 |
+----|-------|-------|----------|---------------+
2 rows in set (0.00 sec)
SELECT id, CONCAT(fname, ' ', lname) AS name
FROM employees PARTITION (p0) ORDER BY lname;
+----|----------------+
| id | name |
+----|----------------+
| 3 | Ellen Johnson |
| 4 | Jim Smith |
| 1 | Bob Taylor |
| 2 | Frank Williams |
+----|----------------+
4 rows in set (0.06 sec)
SELECT store_id, COUNT(department_id) AS c
FROM employees PARTITION (p1,p2,p3)
GROUP BY store_id HAVING c > 4;
+---|----------+
| c | store_id |
+---|----------+
| 5 | 2 |
| 5 | 3 |
+---|----------+
2 rows in set (0.00 sec)
Partition selection is supported for all types of table partitioning, including range partitioning and hash partitioning. For hash partitions, if partition names are not specified, p0
, p1
, p2
,..., or pN-1
is automatically used as the partition name.
SELECT
in INSERT ... SELECT
can also use partition selection.
Restrictions and limitations on partitions
This section introduces some restrictions and limitations on partitioned tables in TiDB.
Partitioning keys, primary keys and unique keys
This section discusses the relationship of partitioning keys with primary keys and unique keys. The rule governing this relationship can be expressed as follows: Every unique key on the table must use every column in the table's partitioning expression. This also includes the table's primary key, because it is by definition a unique key.
For example, the following table creation statements are 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 has at least one unique key that does not include all columns used in the partitioning expression.
The valid statements are as follows:
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;
The following example displays an error:
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. After the following modifications, the CREATE TABLE
statement becomes valid:
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 (col1, col3)
)
PARTITION BY HASH(col1 + col3)
PARTITIONS 4;
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,
UNIQUE KEY (col1, col3),
UNIQUE KEY (col2, col4)
);
Because every primary key is by definition a unique key, so 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 the above examples, the primary key does not include all columns referenced in the partitioning expression. After adding the missing column in the primary key, the CREATE TABLE
statement becomes valid:
CREATE TABLE t5 (
col1 INT NOT NULL,
col2 DATE NOT NULL,
col3 INT NOT NULL,
col4 INT NOT NULL,
PRIMARY KEY(col1, col2, col3)
)
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, col2, col3),
UNIQUE KEY(col2)
)
PARTITION BY HASH( YEAR(col2) )
PARTITIONS 4;
If a table has neither unique keys nor primary keys, then this restriction does not apply.
When you change tables using DDL statements, you also need to consider this restriction when adding a unique index. For example, when you create a partitioned table as shown below:
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)
You can add a non-unique index by using ALTER TABLE
statements. But if you want to add a unique index, the c1
column must be included in the unique index.
Partitioning limitations relating to functions
Only the functions shown in the following list are allowed in partitioning expressions:
ABS()
CEILING() (see CEILING() and FLOOR())
DATEDIFF()
DAY()
DAYOFMONTH()
DAYOFWEEK()
DAYOFYEAR()
EXTRACT() (see EXTRACT() function with WEEK specifier)
FLOOR() (see CEILING() and FLOOR())
HOUR()
MICROSECOND()
MINUTE()
MOD()
MONTH()
QUARTER()
SECOND()
TIME_TO_SEC()
TO_DAYS()
TO_SECONDS()
UNIX_TIMESTAMP() (with TIMESTAMP columns)
WEEKDAY()
YEAR()
YEARWEEK()
Compatibility with MySQL
Currently, TiDB only supports range partitioning and hash partitioning. Other partitioning types that are available in MySQL such as list partitioning and key partitioning are not supported yet in TiDB.
For a table partitioned by RANGE COLUMNS
, currently TiDB only supports using a single partitioning column.
With regard to partition management, any operation that requires moving data in the bottom implementation is not supported currently, including but not limited to: adjust the number of partitions in a hash partitioned table, modify the range of a range partitioned table, merge partitions and exchange partitions.
For the unsupported partitioning types, when you create a table in TiDB, the partitioning information is ignored and the table is created in the regular form with a warning reported. INFORMATION_SCHEMA.PARTITION
tables are not supported currently in TiDB.
The LOAD DATA
syntax does not support partition selection currently in TiDB.
create table t (id int, val int) partition by hash(id) partitions 4;
The regular LOAD DATA
operation is supported:
load local data infile "xxx" into t ...
But Load Data
does not support partition selection:
load local data infile "xxx" into t partition (p1)...
For a partitioned table, the result returned by select * from t
is unordered between the partitions. This is different from the result in MySQL, which is ordered between the partitions but unordered inside the partitions.
create table t (id int, val int) partition by range (id) (
partition p0 values less than (3),
partition p1 values less than (7),
partition p2 values less than (11));
Query OK, 0 rows affected (0.10 sec)
insert into t values (1, 2), (3, 4),(5, 6),(7,8),(9,10);
Query OK, 5 rows affected (0.01 sec)
Records: 5 Duplicates: 0 Warnings: 0
TiDB returns a different result every time, for example:
select * from t;
+------|------+
| id | val |
+------|------+
| 7 | 8 |
| 9 | 10 |
| 1 | 2 |
| 3 | 4 |
| 5 | 6 |
+------|------+
5 rows in set (0.00 sec)
The result returned in MySQL:
select * from t;
+------|------+
| id | val |
+------|------+
| 1 | 2 |
| 3 | 4 |
| 5 | 6 |
| 7 | 8 |
| 9 | 10 |
+------|------+
5 rows in set (0.00 sec)