stupid error... can't think today HELP

phence

I am getting this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/www/path/lst1.php on line 45

From this code:

<?	
$sql = "SELECT * FROM catagories WHERE Parent = 'NULL'";
	$q = mysql_query($sql);
	echo "<SELECT name=\"catagories\" size=\"20\" multiple>\n";

// $catagory is an array, and each key in the array is the name of a column.
// stored under each key is the value for that column in the database.
while ($catagory = mysql_fetch_array($q)) {
	echo "<option name=\"" . $catagory['ID'] . "\">" . $catagory['Name'] . "</option>\n";
}
echo ("</select>");
?>

If anyone could help me figure this out, I'd greatly appreciate it!
Thanks! it's just one of those days

jpmoriarty

yeah your query's probbaly not returning any results.

try echoing out mysql_num_rows($q) - my guess is that it will be 0, so when you try to get the array it's telling you to get stuffed!

NoFear

I would call mysql_num_rows($q) and echo it out right after the query. If you're trying to call mysql_fetch_row() and nothing was returned from MySQL in your query, you're going to get the error.

See if you're getting any results, then report back =)

NoFear

You're too quick, jpmoriarty!

rupertbj

I usually get that error when there is either no table created in the database, or when I don't ask for the right table through misspelling. Have a look at your DB and check what the table you're referring to is called (and that it exists) then make sure you have referred to it correctly.

phence

thanks guys... i misspelled categoires**

thanks alot!!!!

cgraz

in the future, you could prevent a simple error like this by using mysql_error()

$q = mysql_query($sql) or die(mysql_error());

Would've told you that the table catagories didn't exist.

Cgraz