<? $query = "SELECT * from mytable"; $result = mysql_db_query("db", $query);
while ($myarray = mysql_fetch_array($result)) { $title = $myarray[title]; echo "$title<br>"; } ?> Can someone PLEASE tell me why the coding above gives the following error: Warning: Supplied argument is not a valid MySQL result resource in /var/web/somesite.com/html/index.php on line 5 It's absurd. It's driving me nuts. I'm about to introduce my computer to the pavement 40 feet below. Thanks in advance for whoever sees what I am sure is a glaring and obvious flaw in the coding. I've been looking at it for an hours and just can't get anything from where I'm standing, maybe a different perpective will help. _________________________________________________________________ Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php