mariadb/mysql-test/suite/vcol/inc/vcol_partition.inc
2016-12-12 20:35:50 +01:00

139 lines
5 KiB
PHP

################################################################################
# inc/vcol_partition.inc #
# #
# Purpose: #
# Testing partitioning tables with virtual columns. #
# #
# #
# #
#------------------------------------------------------------------------------#
# Original Author: Andrey Zhakov #
# Original Date: 2008-09-04 #
# Change Author: Oleksandr Byelkin (Monty program Ab)
# Date: 2009-03-24
# Change: Syntax changed
################################################################################
--source include/have_partition.inc
--disable_warnings
drop table if exists t1;
--enable_warnings
--echo # Case 1. Partitioning by RANGE based on a non-stored virtual column.
CREATE TABLE t1 (
a DATE NOT NULL,
b int as (year(a))
)
PARTITION BY RANGE( b ) (
PARTITION p0 VALUES LESS THAN (2006),
PARTITION p2 VALUES LESS THAN (2008)
);
insert into t1 values ('2006-01-01',default);
insert into t1 values ('2007-01-01',default);
insert into t1 values ('2005-01-01',default);
insert into t1 (a) values ('2007-01-02');
select * from t1;
select partition_name,table_rows,data_length from information_schema.partitions where table_name = 't1';
--echo # Modify the expression of virtual column b
ALTER TABLE t1 modify b int as (year(a)-1);
select * from t1;
select partition_name,table_rows,data_length from information_schema.partitions where table_name = 't1';
drop table t1;
--echo # Case 2. Partitioning by LIST based on a stored virtual column.
CREATE TABLE t1 (a int, b int as (a % 3 ) persistent)
PARTITION BY LIST (a+1)
(PARTITION p1 VALUES IN (1), PARTITION p2 VALUES IN (2));
insert into t1 values (1,default);
select partition_name,table_rows,data_length from information_schema.partitions where table_name = 't1';
select * from t1;
#
# NOTE: The following tests are currently failing due to a
# [suspected] bug in the existing partition functionality.
# Here is what was observed when using mysqld compiled prior
# to adding the virtual column functionality.
# mysql> create table t1 (a int) partition by list (a)
# (partition p1 values in (1), partition p2 values in (2));
# Query OK, 0 rows affected (0.00 sec)
#
# mysql> insert into t1 values (1), (1), (2);
# Query OK, 3 rows affected (0.00 sec)
# Records: 3 Duplicates: 0 Warnings: 0
#
# mysql> select * from t1;
# +------+
# | a |
# +------+
# | 1 |
# | 1 |
# | 2 |
# +------+
# 3 rows in set (0.00 sec)
#
# mysql> alter table t1 reorganize partition p1 into
# (partition p1 values in (3));
# Query OK, 2 rows affected (3.90 sec)
# Records: 2 Duplicates: 2 Warnings: 0
#
# mysql> select * from t1;
# +------+
# | a |
# +------+
# | 2 | <- Two row have been lost!!!
# +------+
# 1 row in set (0.00 sec)
#
#alter table t1 change b b int as ((a % 3)+1) persistent;
#--error ER_NO_PARTITION_FOR_GIVEN_VALUE
#alter table t1 change b b int as (a % 2) persistent;
#select partition_name,table_rows,data_length from information_schema.partitions where table_name = 't1';
select * from t1;
drop table t1;
--echo # Case 3. Partitioning by HASH based on a non-stored virtual column.
CREATE TABLE t1 (
a DATE NOT NULL,
b int as (year(a))
)
PARTITION BY HASH( b % 3 ) PARTITIONS 3;
insert into t1 values ('2005-01-01',default);
insert into t1 values ('2006-01-01',default);
select * from t1;
select partition_name,table_rows,data_length from information_schema.partitions where table_name = 't1';
--echo # Modify the expression of virtual column b
ALTER TABLE t1 modify b int as (year(a)-1);
select * from t1;
select partition_name,table_rows,data_length from information_schema.partitions where table_name = 't1';
drop table t1;
#
# Restrictions when partitioned
#
--error ER_WRONG_EXPR_IN_PARTITION_FUNC_ERROR
create table t1 (a int, b datetime as (now())) partition by hash(b+1) partitions 3;
--error ER_WRONG_EXPR_IN_PARTITION_FUNC_ERROR
create table t1 (a int, b varchar(100) as (user())) partition by hash(b+1) partitions 3;
--error ER_WRONG_EXPR_IN_PARTITION_FUNC_ERROR
create table t1 (a int, b double as (rand())) partition by hash(b+1) partitions 3;