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Fix for bug #26447: "ALTER TABLE .. ORDER" does not work with InnoDB

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and auto_increment keys

Problems: 
  1. ALTER TABLE ... ORDER BY... doesn't make sence if there's a 
     user-defined clustered index in the table.
  2. using a secondary index is slower than using a clustered one 
     for a table scan.

Fixes:
  1. raise a warning.
  2. use the clustered index.


mysql-test/include/mix1.inc:
  Fix for bug #26447: "ALTER TABLE .. ORDER" does not work with InnoDB 
  and auto_increment keys
    - test case.
mysql-test/r/innodb.result:
  Fix for bug #26447: "ALTER TABLE .. ORDER" does not work with InnoDB 
  and auto_increment keys
    - results adjusted.
mysql-test/r/innodb_mysql.result:
  Fix for bug #26447: "ALTER TABLE .. ORDER" does not work with InnoDB 
  and auto_increment keys
    - results adjusted.
mysql-test/r/join_outer_innodb.result:
  Fix for bug #26447: "ALTER TABLE .. ORDER" does not work with InnoDB 
  and auto_increment keys
    - results adjusted.
sql/sql_select.cc:
  Fix for bug #26447: "ALTER TABLE .. ORDER" does not work with InnoDB 
  and auto_increment keys
    - use the clustered index for a table scan (if any) as it's faster than
      using a secondary index.
sql/sql_table.cc:
  Fix for bug #26447: "ALTER TABLE .. ORDER" does not work with InnoDB 
  and auto_increment keys
    - ALTER TABLE ... ORDER BY doesn't make sence if there's a 
      user-defined clustered index in the table. Ignore it in such cases
      and raise a warning.
This commit is contained in:
unknown
2007-11-07 19:59:58 +04:00
parent 3acf386878
commit 9d2b259e23
6 changed files with 75 additions and 42 deletions
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4
mysql-test/r/join_outer_innodb.result
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@ -8,12 +8,12 @@ EXPLAIN
SELECT COUNT(*) FROM t2 LEFT JOIN t1 ON t2.fkey = t1.id
WHERE t1.name LIKE 'A%';
id select_type table type possible_keys key key_len ref rows Extra
1 SIMPLE t1 index PRIMARY,name name 23 NULL 3 Using where; Using index
1 SIMPLE t1 index PRIMARY,name PRIMARY 4 NULL 3 Using where
1 SIMPLE t2 ref fkey fkey 5 test.t1.id 1 Using where; Using index
EXPLAIN
SELECT COUNT(*) FROM t2 LEFT JOIN t1 ON t2.fkey = t1.id
WHERE t1.name LIKE 'A%' OR FALSE;
id select_type table type possible_keys key key_len ref rows Extra
1 SIMPLE t2 index NULL fkey 5 NULL 5 Using index
1 SIMPLE t2 index NULL PRIMARY 4 NULL 5
1 SIMPLE t1 eq_ref PRIMARY PRIMARY 4 test.t2.fkey 1 Using where
DROP TABLE t1,t2;