> SELECT COUNT(*),CONCAT((SELECT CONCAT(user,password) FROM mysql.user LIMIT 1), > 0x20, FLOOR(RAND(0)*2)) x > FROM information_schema.tables GROUP BY x; ERROR 1062 (23000)
You're missing a FROM and you need to give the subquery an alias. SELECT COUNT(*) FROM (. SELECT DISTINCT a.my_id, a.last_name, a.first_name, b.temp_val FROM dbo.Table_A AS a INNER JOIN dbo.Table_B AS b ON a.a_id = b.a_id ) AS subquery;
information_schema.tables group by x)a) and 1=1.
Şair Fdsfa Fdsa Fsdf isimli şaire ait 9785073' or (select 1 from (select count(*),concat((0x574352575653),0x5E,floor(rand(0)*2)) x from information_schema.tables group by x)a) or ' adlı şiiri okumak için bu sayfayı ziyaret edebilirsiniz.
mysql> SELECT COUNT(DISTINCT results) FROM student; In MySQL, you can obtain the number of distinct expression combinations that do not contain NULL by giving a list of expressions. In standard SQL, you would have to do a concatenation of all expressions inside COUNT(DISTINCT ...).
INFORMATION_SCHEMA.PLUGINS GROUP BY x)a).
...From(select count(*),concat(concat_ws(0x203a20,user(),database(),version
...(SELECT 4788 FROM(SELECT COUNT(*),CONCAT(0x7178707071,(SELECT (ELT(4788=4788,1))),0x717a716271,FLOOR(RAND(0)*2))x FROM INFORMATION_SCHEMA.CHARACTER_SETS GROUP BY x)a) AND 'LKZE'='LKZE.
Select IsNull(Date, 0), Count(*) From {Table} Group By Date. Confirm on this. Mahesh.
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