Comparing Multiple Variables when doing a SELECT

poolgod100

Hello I have a simle question, I think.

I am trying to get all results that match three variables submitted through a form. I use the following command.

<?php
$result = mysql_query("SELECT * FROM videos WHERE Artist LIKE '%$Artist%' AND Show LIKE '%$Show%' AND Tape LIKE '%$Tape%'",$db);
while ($myrow = mysql_fetch_array($result) and $i != $display ) {
printf("<tr><td><center>%s</center></td><td><center>%s</center></td><td><center>%s</center></td><td><center>%s</center></td><td><center>%s</center></td><td><center>%s</center></td></tr>\r\n", $myrow["Artist"], $myrow["Show"], $myrow["Year"], $myrow["Length"], $myrow["Quality"], $myrow["Tape"]);
$i++;
}
?>

This works if I remove the Show and Tape search? But get the following error if I dont:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /www/html/videos.php on line 109

Anyone have any ideas?

Thanks

archangel_617b

Just a note that you should place any code you post in [ php ] or [ code ] tags to make it readable.

Originally posted by poolgod100
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /www/html/videos.php on line 109

The best thing to do typically is to print your error so you can trace the problem. Easy to do using [man]mysql_error/man. Usually something like:

$rid = mysql_query($sql, $db);
if (!$rid)
    die(mysql_error());

I'm going to guess though that your table does not have one of the fields 'Tape' or 'Show' if the query works without them.