<?
$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