Re: Problem with MySQL
| From: | Tyler Longren | Date: | Sat, 15 Jun 2002 04:10:41 +0000 |
| Subject: | Re: Problem with MySQL | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-102307@lists.php.net to get a copy of this message | ||
you need to put your $myrow in a while loop:
while ($myrow = mysql_fetch_array($result)) {
$title = $myrow[title];
$videoid = $myrow[videoid];
$catergory = $myrow[catergory];
$appraisal = $myrow[appraisal];
// blah blah blah everything else
}
Tyler Longren
Captain Jack Communications
www.captainjack.com
tyler@captainjack.com
----- Original Message -----
From: "Chuck Payne" <cepayne@magidesign.com>
To: "PHP General" <php-general@lists.php.net>
Sent: Friday, June 14, 2002 11:03 PM
Subject: [PHP] Problem with MySQL
Hi,
I am working on a movie database I have two database that I am calling from
but the problem I am having when I ask it to go and fetch all the movies
with the same title, it stops and only shows one.
Here is a basic layout...
if($videoid) {
$result = mysql_query("SELECT * FROM library WHERE videoid=$videoid",$db);
$myrow = mysql_fetch_array($result);
// The Myrows
$title = $myrow[title];
$videoid = $myrow[videoid];
$catergory = $myrow[catergory];
$appraisal = $myrow[appraisal];
// Some where here it's not working.....
$sql = "SELECT concat_ws(' ', fname, lname)as actor FROM actormovie WHERE
title = '$title' ORDER by lname";
$result = mysql_query($sql);
print $sql;
$actor = "";
while ($myrow = mysql_fetch_array($result)) {
$actor = $myrow[actor];
$actor .= "<A HREF=''>" . $actor .
"</A><BR>\n";
}
What am I doing wrong? It only show one record and it show more.
Chuck Payne