06.06.2012, 15:04
Try using a double percentage sign. If you're using format(), you need to escape the percentage sign properly or it'll cause problems.
A single percentage sign in format() is considered as a placeholder, more than 1 will actually show in the string properly.
A single percentage sign in format() is considered as a placeholder, more than 1 will actually show in the string properly.
Код:
SELECT * FROM `help` WHERE `searchWord1` LIKE '%%%s%%' LIMIT 0 , 30