Warning: Mysql_result(): Supplied Argument Is Not A Valid Mysql Result Resource In ...

Feelay

Feb 1 2008, 08:50 PM

Hey. I am making a "Version 2.0" For my attack script, but I can't make it work.

This is the error I am gettin:

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in

And here is the code:

CODE

$dbQueryHealth = mysql_query("SELECT temphealth FROM characters WHERE user =". $_POST['atkuser']."");
        $currentHealth = mysql_result($dbQueryHealth, 0);

        $dbQueryExp = mysql_query("SELECT exp FROM characters WHERE user = ".$_POST['atkuser']."");
        $currentExp = mysql_result($dbQueryExp, 0);

I have checked the PHP Manual, but I did not find anything there :S
Thanks For All you help
//Feelay

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sparkx

Feb 2 2008, 09:51 PM

It looks like your mysql_query is wrong. Correct me if I am wrong but I am pretty sure your use WHERE Column='$var' not WHERE Column=".$var." Also you seem to have a odd way of result. I usually do it quite differently but I am not sure what the major difference is.
Example

CODE

$atkuser=$_POST['atkuser'];
$row = mysql_fetch_array(mysql_query("SELECT * FROM database
WHERE User='$atkuser'") or die(mysql_error()));
$EXP=$row['EXP'];

After looking at your code a bit I realized that it seem very strange. I have never seen anyone $_Post a variable directly to a Mysql. First off it is EXTRAMLY insecure and secondly anyone that know PHP could hack it. My example above is also hack-able but you can solve that with a simple preg_replace or preg_match. I am also not quite sure why you have , 0) after your result.

Hope this helps at least a little,
Sparkx

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Miles

Feb 3 2008, 11:49 AM

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in [file] is caused by an sql query that isn't done correctly, usually. To fix it I suggest doing what will be the SQL query firstly in a variable then using mysql_query($sqlvariable), that should solve your problem.

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Feelay

Feb 3 2008, 11:59 AM

I did that

Thanks

It is working now.. but as always, I am having a new problem

The value in the database is not updating :S
here the code for the update, should'nt it work =?

CODE

$SeUs = $_SESSION['user'];
$PoUS = $_POST['atkuser'];
mysql_query ("UPDATE characters SET temphealth =\"{$currentHealthYou}\" WHERE user =\"{$SeUs}\"");
mysql_query ("UPDATE characters SET temphealth = \"{$currentHealthEnemy}\" WHERE user =\"{$PoUs}\"");

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TavoxPeru

Feb 8 2008, 05:43 AM

QUOTE(Feelay @ Feb 3 2008, 06:59 AM)

I did that

Thanks

It is working now.. but as always, I am having a new problem

The value in the database is not updating :S
here the code for the update, should'nt it work =?

CODE

You can use the session variable directly in your code, instead of escaping your strings simply use a single quote ('), also, to prevent sql injections use the mysql_real_escape function with the data posted to your script.

And personally i always attach a die() function with the mysql_error() function to every query i make to check if every thing works fine.

Try the following:

CODE

<?php
$PoUS = mysql_real_escape_string($_POST['atkuser']);
mysql_query ("UPDATE characters SET temphealth='$currentHealthYou' WHERE user='" . $_SESSION['user'] ."'") or die(mysql_error() );
mysql_query ("UPDATE characters SET temphealth = '$currentHealthEnemy' WHERE user ='$PoUs'") or die(mysql_error());
?>

Best regards,

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Warning: Mysql_result(): Supplied Argument Is Not A Valid Mysql Result Resource In ... - This Is for My attack Script.

Warning: Mysql_result(): Supplied Argument Is Not A Valid Mysql Result Resource In ... - This Is for My attack Script.