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Bug #14049391 INNODB MISCALCULATES AUTO-INCREMENT AFTER DECREASING

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AUTO_INCREMENT_INCREMENT
Problem:
=======
When auto_increment_increment system variable decreases,
immediate next value of auto increment column is not affected.

Solution:
========
	Get the previous inserted value of auto increment column by
subtracting the previous auto_increment_increment from next
auto increment value. After that calculate the current autoinc value
using newly changed auto_increment_increment variable.

	Approved by Sunny [rb#4394]
This commit is contained in:
Thirunarayanan B
2014-02-10 10:13:35 +05:30
parent 5ffe40022b
commit 7acdf29cb4
3 changed files with 56 additions and 0 deletions
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27
mysql-test/suite/innodb/r/innodb-autoinc.result
View File

@ -1357,3 +1357,30 @@ t1 CREATE TABLE `t1` (
INSERT INTO t1 VALUES ();
ERROR HY000: Failed to read auto-increment value from storage engine
DROP TABLE t1;
#
# Bug #14049391 INNODB MISCALCULATES AUTO-INCREMENT
# AFTER CHANGING AUTO_INCREMENT_INCREMEMENT
#
CREATE TABLE t ( i INT AUTO_INCREMENT, KEY(i) ) ENGINE=InnoDB;
SET auto_increment_increment = 300;
INSERT INTO t VALUES (NULL), (NULL);
SHOW CREATE TABLE t;
Table Create Table
t CREATE TABLE `t` (
`i` int(11) NOT NULL AUTO_INCREMENT,
KEY `i` (`i`)
) ENGINE=InnoDB AUTO_INCREMENT=601 DEFAULT CHARSET=latin1
SET auto_increment_increment = 50;
INSERT INTO t VALUES (NULL);
SELECT * FROM t;
i
1
301
351
SHOW CREATE TABLE t;
Table Create Table
t CREATE TABLE `t` (
`i` int(11) NOT NULL AUTO_INCREMENT,
KEY `i` (`i`)
) ENGINE=InnoDB AUTO_INCREMENT=401 DEFAULT CHARSET=latin1
DROP TABLE t;

14
mysql-test/suite/innodb/t/innodb-autoinc.test
View File

@ -685,3 +685,17 @@ SHOW CREATE TABLE t1;
--error 1467
INSERT INTO t1 VALUES ();
DROP TABLE t1;
--echo #
--echo # Bug #14049391 INNODB MISCALCULATES AUTO-INCREMENT
--echo # AFTER CHANGING AUTO_INCREMENT_INCREMEMENT
--echo #
CREATE TABLE t ( i INT AUTO_INCREMENT, KEY(i) ) ENGINE=InnoDB;
SET auto_increment_increment = 300;
INSERT INTO t VALUES (NULL), (NULL);
SHOW CREATE TABLE t;
SET auto_increment_increment = 50;
INSERT INTO t VALUES (NULL);
SELECT * FROM t;
SHOW CREATE TABLE t;
DROP TABLE t;