I am trying to do a fresh install of a game. I keep getting this:
Warning: mysql_close(): supplied argument is not a valid MySQL-Link resource in D:\AppServ\www\xerogame\includes\functions.php on line 100

Below is what like 100 of that file looks like. Not sure why there is a problem or what I need to change. Any suggestions?

// $AdminPage -> Si on est dans la section admin ... faut le dire ...
function display ($page, $title = '', $topnav = true, $metatags = '', $AdminPage = false) {
global $link, $game_config, $debug, $user, $planetrow;

if (!$AdminPage) {
$DisplayPage = StdUserHeader ($title, $metatags);
} else {
$DisplayPage = AdminUserHeader ($title, $metatags);
}

if ($topnav) {
$DisplayPage .= ShowTopNavigationBar( $user, $planetrow );
}
$DisplayPage .= "<center>\n". $page ."\n</center>\n";

$DisplayPage .= StdFooter();
if (isset($link)) {
mysql_close($link);
}

echo $DisplayPage;

If you can't close the MySQL connection stored in $link, this means there is something wrong with it, so probably there's a connection issue somewhere.

I am sorry if these are dumb questions. I know a little about php and mysql but am pretty new to this.

What do you think could be wrong? I know my server name, database name, username, password for connecting. I just don't know what the problem is :-(

In this part of the code, there is explicitly nothing wrong.

Originally Posted by Insensus

In this part of the code, there is explicitly nothing wrong.

Thank you for looking. If you have any other suggestions please let me know. I really appreciate it.

This game has an install file and I've done things like this several times. Not sure what the issue is...

Well, theres a few possibilities for this error. 1 and 2 involve your host - maxing out your connection limits or something in your PHP configuration file (yes, it may be PHP even though MySQL is returning this error).
I would say though the most likely cause is because of a previous error.
Most of the time when a developer is writing a script in every query they check for errors -

$result = mysql_query($query) or die ('Error in query: ' . mysql_error());

But sometimes the programmer is lazy and just runs the query without checking for errors -

$result = mysql_query($query);

It may be a query, it may be a connection issue...

You can leave off the $link part. Just use mysql_close();

Once a connection is made the system know what it is connected to.

You should still be able to pass the connection link though. I think what's going on is that the link is never created. Maybe you should use if($link) rather than if(isset($link)). See if that makes a difference.

You can probably get away without closing the link at all because PHP will do this automatically at the end of execution.