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BUG#11766317: FIND_IN_SET won't work normaly after upgrade
from 5.1 to 5.5 (Former 59405) In this bug, args[0] in an Item_func_find_in_set stored an Item_func_weekday that was constant. In Item_func_find_in_set::fix_length_and_dec(), args[0]->val_str() was called. Later, when Item_func_find_in_set::val_int() was called, args[0]->null_value was checked. However, the Item_func_weekday in args[0] had now been replaced with an Item_cache. No val_*() calls had been made to this Item_cache, thus null_value was incorrectly 'true', resulting in missing rows in the result set. enum_value gets a value in fix_length_and_dec() iff args[0] is both constant and non-null. It is therefore unnecessary to check the null_value of args[0] in val_int(). An alternative fix would be to call args[0]->val_int() inside Item_func_find_in_set::val_int(). This would ensure args[0]->null_value was set correctly (always false in this case), but that would have to be done for every record this const value is checked against.
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@ -159,3 +159,45 @@ SELECT CONVERT( a USING latin1 ) FROM t2;
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CONVERT( a USING latin1 )
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DROP TABLE t1, t2;
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#
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# BUG#59405: FIND_IN_SET won't work normaly after upgrade from 5.1 to 5.5
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#
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CREATE TABLE t1(days set('1','2','3','4','5','6','7'));
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INSERT INTO t1 VALUES('1,2,3,4,5,6,7'), (NULL), ('1,2,3,4,5,6,7');
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SELECT * FROM t1 WHERE FIND_IN_SET(DAYOFWEEK(CURRENT_DATE()), days);
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days
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1,2,3,4,5,6,7
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1,2,3,4,5,6,7
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SELECT * FROM t1 WHERE FIND_IN_SET(DAYOFWEEK(CURRENT_DATE()), days) IS UNKNOWN;
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days
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NULL
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SELECT * FROM t1 WHERE FIND_IN_SET(DAYOFWEEK(CURRENT_DATE()), NULL);
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days
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SELECT * FROM t1 WHERE FIND_IN_SET(DAYOFWEEK(CURRENT_DATE()), NULL) IS UNKNOWN;
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days
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1,2,3,4,5,6,7
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NULL
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1,2,3,4,5,6,7
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SELECT * FROM t1 WHERE FIND_IN_SET(7, days);
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days
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1,2,3,4,5,6,7
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1,2,3,4,5,6,7
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SELECT * FROM t1 WHERE FIND_IN_SET(8, days);
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days
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SELECT * FROM t1 WHERE FIND_IN_SET(NULL, days);
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days
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SELECT * FROM t1 WHERE FIND_IN_SET(NULL, days) IS UNKNOWN;
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days
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1,2,3,4,5,6,7
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NULL
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1,2,3,4,5,6,7
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SELECT * FROM t1 WHERE FIND_IN_SET(NULL, NULL);
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days
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SELECT * FROM t1 WHERE FIND_IN_SET(NULL, NULL) IS UNKNOWN;
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days
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1,2,3,4,5,6,7
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NULL
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1,2,3,4,5,6,7
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DROP TABLE t1;
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