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Subject Author Date
problems with login script morph.1989 05-19-2008
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Posted by morph.1989 on May 19, 2008, 7:36 am
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 Hi, I can't get this script to work.
I've used this exact script on other places and it works, but now i
get this error.

<code> Warning: mysql_fetch_array(): supplied argument is not a valid
MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
line 15 </code>

I can't see what is wrong.
Here is the script.

<code>
<?php
session_start();
$anvnamn = $_POST['usr'];
$losenord = $_POST['pwd'];

include "dbconnect.php";

$anv2 = mysql_real_escape_string($anvnamn, $dbconnect);
$los2 = mysql_real_escape_string($losenord, $dbconnect);

$sqlfraga = "SELECT anvnamn FROM administrator WHERE anvnamn = '" .
                        $anvnamn . "' AND losen = '" . $losenord . "'";
$res = mysql_query($sqlfraga, $dbconnect);

if($rad = mysql_fetch_array($res))
{
        $_SESSION['logged_in_admin'] = true;
}
else
{
        $_SESSION['logged_in_admin'] = false;
}
?>
<html>
<body>
<?php
        if($_SESSION['logged_in_admin'])
        {
                echo("You are logged in");
                include('index.php');

        }
        else
        {
                echo ("go away");
        }
?>
</body>
</html>
</code>

Posted by Sjoerd on May 19, 2008, 7:59 am
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 On May 19, 1:36=A0pm, morph.1...@gmail.com wrote:
> <code> Warning: mysql_fetch_array(): supplied argument is not a valid
> MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
> line 15 </code>
>
> $res =3D mysql_query($sqlfraga, $dbconnect);
> if($rad =3D mysql_fetch_array($res))

When the query fails, mysql_query() returns false, which results in
the error message you wrote. I am not sure if this is the case in your
situation, because this would also print a warning. Check the output
of mysql_query() and use mysql_error() to get the error message.




Posted by Robin on May 19, 2008, 8:05 am
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morph.1989@gmail.com wrote:
> Hi, I can't get this script to work.
> I've used this exact script on other places and it works, but now i
> get this error.
>
> <code> Warning: mysql_fetch_array(): supplied argument is not a valid
> MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
> line 15 </code>
>
> I can't see what is wrong.
> Here is the script.
>
> <code>
> <?php
> session_start();
> $anvnamn = $_POST['usr'];
> $losenord = $_POST['pwd'];
>
> include "dbconnect.php";
>
> $anv2 = mysql_real_escape_string($anvnamn, $dbconnect);
> $los2 = mysql_real_escape_string($losenord, $dbconnect);

You create some escaped versions of the $_POST data...

> $sqlfraga = "SELECT anvnamn FROM administrator WHERE anvnamn = '" .
>                         $anvnamn . "' AND losen = '" . $losenord . "'";

... but then fail to use them (SQL injection alert!).

> $res = mysql_query($sqlfraga, $dbconnect);

Then fail to check whether $res is FALSE, which could be the case if
there was an issue with rights to the database.

> if($rad = mysql_fetch_array($res))

Which would cause this to error as described.

So, the error said that $res wasn't valid, so why didn't you check what
was being used? Simple debugging...

Robin

Posted by Rik Wasmus on May 19, 2008, 8:08 am
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> morph.1989@gmail.com wrote:
>> Hi, I can't get this script to work.
>> I've used this exact script on other places and it works, but now i
>> get this error.
>> <code> Warning: mysql_fetch_array(): supplied argument is not a vali=
d
>> MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
>> line 15 </code>
>> I can't see what is wrong.
>> Here is the script.
>> <code>
>> <?php
>> session_start();
>> $anvnamn =3D $_POST['usr'];
>> $losenord =3D $_POST['pwd'];
>> include "dbconnect.php";
>> $anv2 =3D mysql_real_escape_string($anvnamn, $dbconnect);
>> $los2 =3D mysql_real_escape_string($losenord, $dbconnect);
>
> You create some escaped versions of the $_POST data...
>
>> $sqlfraga =3D "SELECT anvnamn FROM administrator WHERE anvnamn =3D '"=
.
>>                         $anvnamn . "' AND losen =3D '" . $losenord . "'";
>
> ... but then fail to use them (SQL injection alert!).

Hehe, you beat me to it, with even more or less the same format/wording =
:D
-- =

Rik Wasmus
...spamrun finished

Posted by Rik Wasmus on May 19, 2008, 8:07 am
Please log in for more thread options

> Hi, I can't get this script to work.
> I've used this exact script on other places and it works, but now i
> get this error.
>
> <code> Warning: mysql_fetch_array(): supplied argument is not a valid
> MySQL result resource in C:\xampp\htdocs\uploads\login_script.php on
> line 15 </code>
>
> I can't see what is wrong.
> Here is the script.
>
> <code>
> <?php
> session_start();
> $anvnamn =3D $_POST['usr'];
> $losenord =3D $_POST['pwd'];
>
> include "dbconnect.php";
>
> $anv2 =3D mysql_real_escape_string($anvnamn, $dbconnect);
> $los2 =3D mysql_real_escape_string($losenord, $dbconnect);

Proper escaping and then:

> $sqlfraga =3D "SELECT anvnamn FROM administrator WHERE anvnamn =3D '" =
.
>                         $anvnamn . "' AND losen =3D '" . $losenord . "'";

... using the unescaped variables!

You, my friend, are vulnerable to SQL injection. Use the $avn2 & $los2 =

variables in the query, that's why you escape()d them...

If you still have the same problem, echo $sqlfraga & mysql_error() to t=
he =

screen and check what's wrong with the query.
-- =

Rik Wasmus
...spamrun finished

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