Query not running on subsequent page recalls

[PHPSardine]

I have a PHP template system for page page such that the following code will always generate a similar looking page:

<?php

include('header.html');


//Page specific content goes here

include('footer.html');

?>

This system has been working fine, but I just tried adding a dropdown list menu that is generated use values stored in a database. On the initial page load, everything works fine, then when I click go to go to another page, the list will still be there, but it's empty...suggestings that the query isn't running the second time around. Here is the code I am using; it is located in footer.html:

	<?php

	require_once('includes/mysql_connect.php');
	$query = "SELECT LEFT(track, 15) AS name, track_id FROM track ORDER BY track ASC";
	$result = mysql_query($query);

	echo "<form action=\"songs.php\" method=\"get\">
		<select name=\"track\">";

	while($row = mysql_fetch_array($result)){
		if(strlen($row['name']) == 15){
			echo "<option value=\"{$row['track_id']}\">{$row['name']}...</option>";
		} else {
			echo "<option value=\"{$row['track_id']}\">{$row['name']}</option>";
		}
	}
	mysql_free_result($result);

	echo "</select>
		<input type=\"submit\" value=\"Go\"/>
		</form><br/>";
?>

I also tried loading the database info into an array, hoping that the array would spit out the values that I wanted consistently...but that did not work either. Here's the code:

	<?php

	require_once('includes/mysql_connect.php');
	$query = "SELECT LEFT(track, 15) AS name, track_id FROM track ORDER BY track ASC";
	$result = mysql_query($query);

	while($row = mysql_fetch_array($result)){
		if(strlen($row['name']) == 15){
			$track[$row['track_id']] = "{$row['name']}...";
		} else {
			$track[$row['track_id']] = "{$row['name']}";
		}
	}
	echo "<form action=\"songs.php\" method=\"get\">
		<select name=\"track\">";

	foreach($track as $key => $value){
		echo "<option value=\"$key\">$value</option>";
	}

	echo "</select>
		<input type=\"submit\" value=\"Go\"/>
		</form><br/>";
?>

Thanks so much for the help!!!

[devinemke]

where is this code located? in your header file? your form action points to songs.php, is this code called again in that file? you should also use [man]mysql_error[/man] while debugging.

[PHPSardine]

It's located in the footer.html file. I forgot to post that and then edited it. I will try the debugging.

[PHPSardine]

Here is the error message I'm getting:

A fatal MySQL error occured.
Query: SELECT LEFT(track, 15) AS name, track_id FROM track ORDER BY track ASC
Error: (1045) Access denied for user 'nobody'@'ip' (using password: NO)

This is really weird because the query is running fine initially, then when I select a track in the drop down, it'll take me to the correct page and then it won't reload the list (and now that I'm using the error tracking, I get this error). What's even weirder is that I made another drop down just below that one with essentially the same code and it's working perfectly.

[PHPSardine]

I figured it out. I had a mysql_close() function in another page. The connection was closed, so I was getting through....thanks for the help!