PHP: Class not found

Question

I'm getting this annoying error and haven't been able to fix it yet.

<b>Fatal error: Class 'Console' not found in /home/serellyn/public_html/HEIM/php/nieuwbeheer/console_overview.php on line 45.</b>

Let's first start with the hierarchy which is like this.

index (main page) console_overview (section of page) include/connect (connect to DB) include/console.class (the class)

The index.php requires the connect.php and the console.class.php and loads the console_overview.php. Here's the code:

<?php
require_once('include/connect.php');
require_once('include/console.class.php');
var_dump(file_exists('include/connect.php'));
var_dump(file_exists('include/console.class.php'));
?>

<div id="mainpage" class="main-container inner">
    <?php
        if (isset($_GET['page'])) {
            $page = $_GET['page'];
        } else {
            $page = "console_overview";
        }
    ?>
</div>
<!-- end: MAIN CONTAINER -->

<script>
var page = "<?php echo $page;?>";
$( "#mainpage" ).load( page + ".php" );
</script>

I've used var_dumps to check if both file exists (and they do). The console_overview.php loads correctly. Now in the console_overview.php I'm trying to get data from the Console class, as following:

<?php
foreach(Console::getAllConsoles() as $aConsole) {
    $consoleID= $aConsole->getID();
    $consoleName = $aConsole->getName();
    $consoleHostname = $aConsole->getHostname();
    $consoleGameID = $aConsole->getGameID();
    $consolePowerState = $aConsole->getPowerState();

    echo "<tr>";
    echo "1";
}
?>

The error I'm getting is caused by the foreach... but I can't find out what's wrong...

The Console class looks like this (I'm pasting the most important parts, otherwise the code would be too long).

<?php

class Console{

private $ID, $hostname, $mac, $ip, $roomID, $gameID, $register, $powerState, $dateUpdated;  

public function Console($tID, $tHostname, $tMac, $tIp, $tRoomID, $tGameID, $tRegister, $tPowerState, $tDateUpdated) {
    $this->ID = $tID;
    $this->hostname = $tHostname;
    $this->mac = $tMac;
    $this->ip = $tIp;
    $this->roomID = $tRoomID;
    $this->gameID = $tGameID;
    $this->register = $tRegister;
    $this->powerState = $tPowerState;
    $this->dateUpdated= $tDateUpdated;
}
...
public static function getAllConsoles() {
    $sql = "SELECT * FROM `console` ORDER BY `hostname` ASC";
    $result = mysql_query($sql);
    $theResults = array();
    while ($row = mysql_fetch_array($result)) {
        $theResults[] = new      Console($row[0],$row[1],$row[2],$row[3],$row[4],$row[5],$row[6],$row[7],$row[8],$row[9],$row[10]);
    }
    return $theResults;
}

}

?>

So can anyone see what the problem is? Thank you for your help.

Edit: O and yes, I know MySQL is deprecated and will change this whenever the issue of not finding the console is fixed =).

You can't include a file with javascript and expect it to understand what's going on in PHP. These are two seperate PHP calls. Every time you get a PHP site off a website (unless you use require inside PHP) it creates a new request towards the server. You need to include the PHP files in the javascript loaded PHP files. — h2ooooooo
– h2ooooooo, Commented Sep 1, 2014 at 8:04
Side note: The mysql_* functions you are using are becoming deprecated and will be removed from future versions of PHP. You should not use them to write new code. Use mysqli_* or PDO instead. — Gerald Schneider
– Gerald Schneider, Commented Sep 1, 2014 at 8:07
I get it now, and yes Gerald, thanks for the warning. I do know this and is on my to do list ;) — Serellyn
– Serellyn, Commented Sep 1, 2014 at 13:21

Gerald Schneider · Accepted Answer · 2014-09-01 08:18:53Z

4

Your console_overview.php does not include the required files. When you make an AJAX call with JavaScript from the client it is a separate HTTP request to the server, so you have to add the require() call again there:

<?php
require_once('include/connect.php');
require_once('include/console.class.php');
foreach(Console::getAllConsoles() as $aConsole) {
    $consoleID= $aConsole->getID();
    $consoleName = $aConsole->getName();
    $consoleHostname = $aConsole->getHostname();
    $consoleGameID = $aConsole->getGameID();
    $consolePowerState = $aConsole->getPowerState();

    echo "<tr>";
    echo "1";
}
?>

edited Sep 1, 2014 at 8:18

answered Sep 1, 2014 at 8:04

Gerald Schneider

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2 Comments

Serellyn Over a year ago

Ah, right, thank you! One more thing, I'm getting the following error now: mysql_fetch_array() expects parameter 1 to be resource, boolean given. It's from $result = mysql_query($sql); $theResults = array(); while ($row = mysql_fetch_array($result)) { $theResults[] = new Console($row[0],$row[1],$row[2],$row[3],$row[4],$row[5],$row[6],$row[7],$row[8],$row[9],$row[10]); } Any idea why?

Gerald Schneider Over a year ago

echo mysql_error(); is your friend. This question has been answered over and over on SO.

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