Re: Driving me nuts, need one second of your time

It looks like your sql query failed, so the result is invalid. Assuming that this is really the whole script, then you're counting on the mysql_db_query function to open a connection to the database "db" using the default connection values which (according to the manual) are host: localhost, user: (whatever user is running the script, probably 'www', 'apache', or 'nobody' depending on your system), and password: (blank). Do you really have your mysql database setup to allow access with these defaults? I'd recommend trying mysql_connect() first, then if that works continue on to the mysql_query(). At least you'll know better which aspect failed. -Steve On Friday, March 15, 2002, at 05:14 PM, cosmin laslau wrote:

<? $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