The COUNT(*) and GROUP BY parts are necessary to provoke the error in MySQL: > SELECT COUNT(*),FLOOR(RAND(0)*2)x FROM information_schema.tables GROUP BY x; ERROR 1062 (23000): Duplicate entry '1' for key 'group_key'.
Shouldn't select count(*) take more time to execute since it makes more operations? To output all the results from select * I need 4 minutes (not 32 seconds, as indicated by server log). I understand that the client has to output a lot of data and it will be slow, but what about the server's log?
SQL> select x, count(*) from t01 group by x; no rows selected. try to remove the group by to see the difference.
...5.0 AND error-based - WHERE, HAVING, ORDER BY or GROUP BY clause (FLOOR) # Payload: id=5' AND (SELECT 7446 FROM(SELECT COUNT(*),CONCAT(0x7178707871,(SELECT (ELT
The boot2root is called NullByte 0x01 and is described as beginner/intermediate level challenge. I thought it was pretty easy, but still a fun challenge
Login to your bWAPP and select vulnerability SQL Injection (Login Form/Hero). As stated in previous post, we need to do some
data: lv_extractrows TYPE i. Select count(*) from (p_tabname) bypassing buffer. WHERE (cond_syntax).
select Count(*) as Num_Rows from SysObjects where Type = 'U' and Name = 'System_Configuration'.
If you name columns to select in addition to the COUNT() value, a GROUP BY clause should be present that names those same columns.
I'm counting the number of reimbursement by year, month and by ATC (a categorical variable describing "classes" of medicines).