不使用include语句传输的PHP变量的值

OK, I know that I am being stupid about this, but I can't seem to figure out how I am being so.

I have variables for social media links in a separate .php file so that my client can change them without having to see too much code.

I include that .php file in my normal code where I want the links to be. But, for some reason the value of the variable is not being transmitted. I even tried to make a "global" variable. Here is my code.

File: socialMediaURLs.php

<?php

    global $facebook = 'https://www.facebook.com/MenCoachingMen'; 
    $googlePlus = 'https://plus.google.com/104275309033865331192/posts';
    $twitter = 'https://twitter.com/MenCoachingMen';
    $rss = 'http://mencoachingmen.org/category/podcast/feed/';
    $vimeo = 'https://vimeo.com/mencoachingmen';
    $youtube = 'https://www.youtube.com/channel/UCy_Pth5x-O7rcX9nMx1e8qw';

?>

File: socialLinks.php

<?php include './socialMediaURLs.php'; ?>

<div class="social_links_wrapper">

    <a href="<?php echo $facebook;?>">
        <div class="sl_facebook">
            <i class="fa fa-facebook fa-5x"></i>
        </div>
    </a>

    <a href="<?php echo $googlePlus;?>">
        <div class="sl_googlePlus">
            <i class="fa fa-google-plus fa-5x"></i>
        </div>
    </a>    

    <a href="<?php echo $twitter;?>">
        <div class="sl_twitter">
            <i class="fa fa-twitter fa-5x"></i>
        </div>
    </a>

    <a href="<?php echo $rss;?>">
        <div class="sl_rss">
            <i class="fa fa-rss fa-5x"></i>
        </div>
    </a>

    <a href="<?php echo $vimeo;?>">
        <div class="sl_vimeo">
            <i class="fa fa-vimeo-square fa-5x"></i>
        </div>
    </a>

    <a href="<?php echo $youtube;?>">
        <div class="sl_youtube">
            <i class="fa fa-youtube fa-5x"></i>
        </div>
    </a>

</div>

I ran a test like so:

<?php 
    if($facebook){
        echo $facebook;
    }else{
        echo 'Facebook NULL';
    }
?>

I put this after the include statement and before the rest of the code. And it prints out 'Facebook NULL'. So, I know that the value is not being transmitted. Now, if I put in the included .php file (where the variables are stored) a echo "Hello World!"; line, "Hello World!" does print out on the screen. So, I know that the file is being included correctly (ie the path is correct).

I then placed this code after the 'include' and before the rest of the code:

$facebook = 'https://www.facebook.com/MenCoachingMen'; 

When I do that, the URL is included within the page. So, I know that my php statements within the code is correct. That means that it has to be a transmission of the value of the variable after including it. Please help. I know this must be a stupid mistake somewhere. Thank you.

A global variable must be used inside a function.

function name(){
    global $globalvar = "some info";
}

to pass a variable OUT of it and into the script. You can't pass variables to other scripts using global variables. Consider using $_SESSION['name'] variables to pass variables from script to script.

http://php.net/manual/en/book.session.php

<?php include 'socialMediaURLs.php'; ?> <--- top main public_html/www
<?php include 'directory1/socialMediaURLs.php'; ?> <-- one folder deep
<?php include 'directory1/directory2/socialMediaURLS.php'; ?> <-- two folders deep

//calling a file FROM two folders deep TO A FILE IN the root folder!    
<?php include('../../socialMediaURLS.php'; ?>  <--- main index public_html/www
//ETC......
<?php include('../../directory1/socialMediaURLS.php'; ?>

I hope this helps. Think of it like DIR and CD commands.