Problem with PHP/MySQL - error message on line 13

whitestream6 · Joined: 14 Dec 2009 Posts: 3

I get this error in my PHP script (which is included as a file within a main PHP file):
[quote]Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\www\vhosts\myradiostation.co.uk\showfile.php on line 13[/quote]

This is the full code of the script (line 13 is highlighted in bold text):
[quote] <?php
//connect to mysql
//change user and password to your mySQL name and password
mysql_connect("localhost","root","PASSWORDREMOVED");

//select which database you want to edit
mysql_select_db("myradio");

//select the table
$result = mysql_query("select * from presenters");

//grab all the content
[b]while($r=mysql_fetch_array($result)) [/b]
//the format is $variable = $r["nameofmysqlcolumn"];
//modify these to match your mysql table columns
{
$shows=$r["shows"];
$onair=$r["onair];
$images=$r["image"];
$showdesc=$r["showdesc"];
echo "<div class=\"divider\"></div>
<div class=\"main\" style=\"width:552px;\">
<img src=\"$images\" width=115 height=60>
<div class=\"time\">$onair</div>
<div class=\"show\"><h3><b>$shows</b></h3>
<p>$showdesc</p></div>
<div class=\"footer\"></div>
</div>";
}
?>[/quote]

How do I fix this error myself - I have tried to look it up on Google and fix it myself, but so far I haven't had much luck!

I'm running Apache on localhost, if that's of any interest, with PHP 5.25.

lostboy · Joined: 02 May 2004 Posts: 6058 Location: toronto, canada

Its usually an error in the query or connection. Try this

[code:1]
$result = mysql_query("select * from presenters") or die(mysql_error());
[/code:1]