Is there a way to put this PHP into an array and simplify it?

Question

The following code loads all .php files found in the specified folder (defined separately). Is there a way to put this into an array to simplify the code?

Only a couple of variables change but essentially the code repeats several times.

// The General Files

$the_general = opendir(FRAMEWORK_GENERAL);

while (($the_general_files = readdir($the_general)) !== false) {
    if(strpos($the_general_files,'.php')) {
        include_once(FRAMEWORK_GENERAL . $the_general_files);       
    }       
}

closedir($the_general);


// The Plugin Files

$the_plugins = opendir(FRAMEWORK_PLUGINS);

while (($the_plugins_files = readdir($the_plugins)) !== false) {
    if(strpos($the_plugins_files,'.php')) {
        include_once(FRAMEWORK_PLUGINS . $the_plugins_files);       
    }       
}

closedir($the_plugins);

There are several more sections which call different folders.

Any help is greatly appreciated.

Cheers, James

Answer 1

I nicer way to do this would to use glob(). And make it into a function.

function includeAllInDirectory($directory)
{
    if (!is_dir($directory)) {
        return false;
    }

    // Make sure to add a trailing slash
    $directory = rtrim($directory, '/\\') . '/';

    foreach (glob("{$directory}*.php") as $filename) {
        require_once($directory . $filename);
    }

    return true;
}

Answer 2

This is fairly simple. See arrays and foreach.

$dirs = array(FRAMEWORK_GENERAL, FRAMEWORK_PLUGINS, );

foreach ($dirs as $dir) {
    $d = opendir($dir);

    while (($file = readdir($d)) !== false) {
        if(strpos($file,'.php')) {
            include_once($dir . $file);       
        }       
    }

    closedir($d);
}