[SOLVED] Error in code

[Ell20]

Hey,

 

Im using part of some code I created on a previous page to help me with a new page im creating. I have changed around the variables so they respond to the correct page however I am getting an error which I dont understand:

 

An error occured in script c:\program files\easyphp1-8\www\html\events.php on line 43: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource

 

The code works on the previous page but dosent work on this one, when all I changed is the variables!!

 

$query = ("SELECT * FROM event WHERE club_id = '$club_id' ORDER BY 'date' DESC")
OR DIE(mysql_error());
$result = mysql_query($query);

//Line 43
while ($row = mysql_fetch_assoc($result)) {
$eventid = $row['event_id'];

 

Cheers for any help

 

Elliot

[AtomicNetwork]

try replacing the following line:

 

$query = ("SELECT * FROM event WHERE club_id = '$club_id' ORDER BY 'date' DESC")

 

with

 

$query = "SELECT * FROM event WHERE club_id = '$club_id' ORDER BY `date` DESC";

[MadTechie]

try this

Read comments

<?php
//Check your connecting to the database.. (i assume thats okay)

//moved the or die to the query (date is a field so use backtick)
$query = "SELECT * FROM event WHERE club_id = '$club_id' ORDER BY `date` DESC";
$result = mysql_query($query)or die(mysql_error());

//Line 43
while ($row = mysql_fetch_assoc($result))
{
$eventid = $row['event_id'];
?>

[Ell20]

Thanks MadTechie, your code produced a mysport.event table does not exist!

 

It was a typo and should have read .......FROM events WHERE.......

 

Thanks for the help