Hey guys, I'm fairly new to php and so I need some help on some stuff that should be basic. I'm trying to implement the following code to grab the name of a video, that I have stored in a mysql database. The video name is in the same row as the variable $url. When I try to implement the code I get the following warning "Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource" And the warning refers to me to the line where I have put ####. This might seem like a rookie mistake but any help would be appreciated.

PHP Code:


<?php 
$url = ['www.mysite.com/examplevideo'];
 ?>

<?php
$query = mysql_query("SELECT video_name FROM videos WHERE video_url = '$url' ");
while ($row = mysql_fetch_assoc($query))   ####
 {
$videoname = $row['video_name'];
}    
?>

It means your squery failed to return any results.

So there wouldn't be any errors in my code then? its just a matter of the query returning nothing.

So there wouldn't be any errors in my code then

Who knows?
We can only see a small part of the code.

Add some error checking to to the queries

http://www.php.net/manual/en/function.mysql-query.php

hmm thanks, ill look into the link.

PHP Code:

$query = mysql_query("SELECT video_name FROM videos WHERE video_url = '$url' ");
while ($row = mysql_fetch_assoc($query)); 


You have one too many " ) " ?

PHP Code:

$query = mysql_query("SELECT video_name FROM videos WHERE video_url = '$url' ");
while ($row = mysql_fetch_assoc($query); 


It said that you query is not good.