mysql fetch array error driving me batty!

[foreverhex]

I don't know if its because Im tired or something but I seem to be over looking something. Could some one please help... I think I just lost all sanity. I keep getting

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in ...shop.php on line 102

 

This is a normal error, so I go back to see if somethings wrong with my coding in SQL and I can find it. Here is the code

 

<?php
//There is other crap before this that in theory shouldn't matter.

	elseif ($_GET[action] == 'close') {
		//First get the order and make it nice and viewable
		include 'rhf_db.php';
		mysql_select_db('rhf_orders') or die('Could not select database');
		$dbnum = $_SESSION[number];
		$sql_getord = ("SELECT * FROM $dbnum WHERE status = 'open' ORDER BY date DESC") or die("Error getting order: " . mysql_error());

		$sqlnum = '0';
		while ($row_getorder = mysql_fetch_array($sql_getord)) {
			$preorder = $preorder . $row_getorder[item][$sqlnum] . ' | ' . $row_getorder[quantity][$sqlnum] . '<br>';
			$sqlnum++;
		}

		echo 'preorder: <br><br>' . $preorder;
?>

 

Thanks in advance!

[zenag]

 

try this...

sql_getord = mysql_query("SELECT * FROM $dbnum WHERE status = 'open' ORDER BY date DESC")

[benjaminbeazy]

$sql_getord = mysql_query("SELECT * FROM $dbnum WHERE status = 'open' ORDER BY date DESC") or die("Error getting order: " . mysql_error());

[foreverhex]

OMG! all apologizes... I am so tired, wow, I have been programing with php for 4 years and I forget and then over look that. wow I am so ashamed.