| Bug #11745 | insert where not exists failing | ||
|---|---|---|---|
| Submitted: | 5 Jul 2005 20:16 | Modified: | 3 Aug 2005 20:56 |
| Reporter: | Joe Bob | Email Updates: | |
| Status: | Closed | Impact on me: | |
| Category: | MySQL Server: Optimizer | Severity: | S2 (Serious) |
| Version: | 4.1.12a | OS: | Any (All) |
| Assigned to: | Igor Babaev | CPU Architecture: | Any |
[5 Jul 2005 20:16]
Joe Bob
[12 Jul 2005 4:40]
Jorge del Conde
ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE NOT EXISTS (SELECT * FROM SUBSCRIPTIONS WHERE ARTICLE_ID = 50 AND USER_ID ' at line 3 mysql> exit jorge-include/mysql> grep 1064 * mysqld_error.h:#define ER_PARSE_ERROR 1064
[17 Jul 2005 16:46]
Bugs System
A patch for this bug has been committed. After review, it may be pushed to the relevant source trees for release in the next version. You can access the patch from: http://lists.mysql.com/internals/27233
[23 Jul 2005 17:01]
Igor Babaev
Where clause was not accepted with FROM DUAL by our syntax before.
I corrected the grammar rule and added a support for this case.
ChangeSet
1.2337 05/07/17 09:46:14 igor@rurik.mysql.com +4 -0
select.result, select.test:
Added a test case for bug #11745.
sql_select.cc:
Fixed bug # 11745.
Added support of where clause for queries with FROM DUAL.
sql_yacc.yy:
Fixed bug # 11745.
Added optional where clause for queries with FROM DUAL.
The fix will appear in 4.1.14 and 5.0.11
[3 Aug 2005 20:56]
Mike Hillyer
Documented in 4.1.14 and 5.0.11 changelogs: <listitem><para>Added support of where clause for queries with <literal>FROM DUAL</literal>. (Bug #11745)</para></listitem>
