PHP function use variable from outside

Question

function parts  part          structure    http        site url    content         echo  tructure    part     php         This function uses a variable  site url that was defined at the top of this page  but this variable is not being passed into the function    How do we get it to return in the function

User · Answer

I suppose this depends on your architecture and whatever else you may need to consider, but you could also take the object-oriented approach and use a class.

class ClassName {

    private $site_url;

    function __construct( $url ) {
        $this->site_url = $url;
    }

    public function parts( string $part ) {
        echo 'http://' . $this->site_url . 'content/' . $part . '.php';
    }

    # You could build a bunch of other things here
    # too and still have access to $this->site_url.
}

Then you can create and use the object wherever you'd like.

$obj = new ClassName($site_url);
$obj->parts('part_argument');

This could be overkill for what OP was specifically trying to achieve, but it's at least an option I wanted to put on the table for newcomers since nobody mentioned it yet.

The advantage here is scalability and containment. For example, if you find yourself needing to pass the same variables as references to multiple functions for the sake of a common task, that could be an indicator that a class is in order.

User · Answer

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

User · Answer

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    Suggested reading    Variable scope in PHP Anonymous functions

User · Answer

Do not forget that you also can pass these use variables by reference   The use cases are when you need to change the use d variable from inside of your callback  e g  produce the new array of different objects from some source array of objects      sourcearray      object    a    gt  1    object    a    gt  2     newarray       array walk  sourcearray  function   item  use   amp  newarray         newarray      object    times2    gt   item- gt a   2       var dump  newarray     Now  newarray will comprise  pseudocode here for brevity    times2 2   times2 4     On the contrary  using  newarray with no  amp  modifier would make outer  newarray variable be read-only accessible from within the closure scope  But  newarray within closure scope would be a completelly different newly created variable living only within the closure scope    Despite both variables  names are the same these would be two different variables  The outer  newarray variable would comprise    in this case after the code has finishes

User · Answer

Just put in the function using GLOBAL keyword    global  site url

[php] PHP function use variable from outside

Examples related to php

Examples related to function

Examples related to variables