PHP function use variable from outside

Alternatively, you can bring variables in from the outside scope by using closures with the use keyword.

$myVar = "foo";
$myFunction = function($arg1, $arg2) use ($myVar)
{
 return $arg1 . $myVar . $arg2;
};

Add second parameter

You need to pass additional parameter to your function:

function parts($site_url, $part) { 
    $structure = 'http://' . $site_url . 'content/'; 
    echo $structure . $part . '.php'; 
}

In case of closures

If you'd rather use closures then you can import variable to the current scope (the use keyword):

$parts = function($part) use ($site_url) { 
    $structure = 'http://' . $site_url . 'content/'; 
    echo $structure . $part . '.php'; 
};

global - a bad practice

This post is frequently read, so something needs to be clarified about global. Using it is considered a bad practice (refer to this and this).

For the completeness sake here is the solution using global:

function parts($part) { 
    global $site_url;
    $structure = 'http://' . $site_url . 'content/'; 
    echo($structure . $part . '.php'); 
}

It works because you have to tell interpreter that you want to use a global variable, now it thinks it's a local variable (within your function).

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