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RE: Simple to exploit SQL Injection ?

Subject: RE: Simple to exploit SQL Injection ?
Date: Mon, 28 Nov 2005 06:00:38 -0800 (PST)

If I understand correctly, it seems, they escape
quotes on the password field. 
Like in PHP, when magic_quotes_gpc is on AND strings
are used as sql params, SQL injection is impossible.
here even -- trick won't work.
You should try to tamper login parameter.
-pilon

--- "Haaland, Vegar Linge"
<Vegar.Linge.Haaland@palantir.no> wrote:

 And you could try using:
' or ''='
As username and password. That will make the querry
look like:
SELECT * FROM users WHERE username = '' or ''='' AND
password = '' or
''=''
(Or anything that always is true;
Some expamples:
You could use: hi' or 'a'='a
This will give you username = 'hi' or 'a'='a'
This will "always" be true (if I real the querry
right :P) , cause 'a'
equals 'a'
And so on.

-----Original Message-----
From: Yousef Syed [mailto:yousef.syed@gmail.com] 
Sent: 28. november 2005 13:20
To: Jason binger
Cc: webappsec@securityfocus.com
Subject: Re: Simple to exploit SQL Injection ?

Hi Jason,
Try the following Password:
' OR 1=1 --

That should give the following SQL:
'SELECT *
FROM users
WHERE username = 'xyz'
AND password = '' OR 1=1 -- '

Since 1 always evaluates to 1, the rest of the SQL
will be ignored and
you should get the result you were expecting. Using
the "--" comment,
will stop anything else after this from being
evaluated. That should
stop you getting any syntax errors.

ys

--
Yousef Syed
"One senior official said the consultancy "doesn't
have the greatest of
reputations among civil servants. They come and
state the bleeding
obvious using Powerpoint"."

On 28/11/05, Jason binger <cisspstudy@yahoo.com>
wrote:
I am reviewing a .Net web application. When
entering xyz for a 
username and ' for a password into a form I
receive the following 
stack trace (extract):

System.Exception: Can't Load DataReader using SQL
string: 'SELECT * FROM users WHERE username =
'xyz'
AND password = '''' -- Unclosed quotation mark
before the character 
string '''. Line 1: Incorrect syntax near '''.

Now I would have thought this would be easy to
exploit, but I can't 
bypass the logon page. xyz is a valid username.
Any ideas?

Cheers




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