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BUG#16590: Optimized does not do right "const" table pre-read

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st_table::const_key_parts member is used in determining if
 certain key has a prefix that is compared to constant(s) in
 the query predicates.
 If there's such prefix the index can be used to get the data
 from the remaining suffix columns in sorted order.
 However if a field is compared to another field from a "const"
 table the const_key_parts is not amended.
 This makes the optimizer unable to detect that the key can be 
 used for sorting and adds an extra filesort.
 Fixed by updating const_key_parts after reading in the "const"
 table.
This commit is contained in:
gkodinov/kgeorge@macbook.gmz 2007-01-22 12:51:21 +02:00
parent 814c1308ea
commit 9ea140d13e
3 changed files with 38 additions and 0 deletions
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9
mysql-test/r/order_by.result
View File

@ -918,3 +918,12 @@ NULL
2
3
DROP TABLE t1,t2,t3,t4;
CREATE TABLE t1 (a INT, b INT, PRIMARY KEY (a), UNIQUE KEY b (b));
INSERT INTO t1 VALUES (1,1),(2,2);
CREATE TABLE t2 (a INT, b INT, KEY a (a,b));
INSERT INTO t2 VALUES (1,1),(1,2),(2,1),(2,2);
EXPLAIN SELECT 1 FROM t1,t2 WHERE t1.b=2 AND t1.a=t2.a ORDER BY t2.b;
id select_type table type possible_keys key key_len ref rows Extra
1 SIMPLE t1 const PRIMARY,b b 5 const 1
1 SIMPLE t2 ref a a 5 const 2 Using where; Using index
DROP TABLE t1,t2;

13
mysql-test/t/order_by.test
View File

@ -625,3 +625,16 @@ SELECT t2.b FROM t1 LEFT JOIN (t2, t3 LEFT JOIN t4 ON t3.a=t4.a)
ON (t1.a=t2.a AND t1.b=t3.b) order by t2.b;
DROP TABLE t1,t2,t3,t4;
#
# BUG#16590: Optimized does not do right "const" table pre-read
#
CREATE TABLE t1 (a INT, b INT, PRIMARY KEY (a), UNIQUE KEY b (b));
INSERT INTO t1 VALUES (1,1),(2,2);
CREATE TABLE t2 (a INT, b INT, KEY a (a,b));
INSERT INTO t2 VALUES (1,1),(1,2),(2,1),(2,2);
EXPLAIN SELECT 1 FROM t1,t2 WHERE t1.b=2 AND t1.a=t2.a ORDER BY t2.b;
DROP TABLE t1,t2;

16
sql/sql_select.cc
View File

@ -7712,6 +7712,22 @@ static void update_const_equal_items(COND *cond, JOIN_TAB *tab)
key_map possible_keys= field->key_start;
possible_keys.intersect(field->table->keys_in_use_for_query);
stat[0].const_keys.merge(possible_keys);
/*
For each field in the multiple equality (for which we know that it
is a constant) we have to find its corresponding key part, and set
that key part in const_key_parts.
*/
if (!possible_keys.is_clear_all())
{
TABLE *tab= field->table;
KEYUSE *use;
for (use= stat->keyuse; use && use->table == tab; use++)
if (possible_keys.is_set(use->key) &&
tab->key_info[use->key].key_part[use->keypart].field ==
field)
tab->const_key_parts[use->key]|= use->keypart_map;
}
}
}
}