PHP call_user_func_array: How to pass all parameters as a single array?

Question

I am using call_user_func_array in an MVC structure to point to a controller and method and pass the parameters of the URL:

// At this point, $url is an array with nothing but parameters

$this->params = $url ? array_values($url) : []; //

call_user_func_array([$this->controller, $this->method], $this->params);

Now when I use this:

class Home extends Controller {

    public function index($param1=null, $param2=null) { // Ugly...

      // .. continue

Is there any way to send all the parameters to the class method as one single array? There's no telling if there are any parameters to begin with, and if there are, there's no telling how many. I don't want to have to do:

public function index($param1=null, $param2=null, $param3=null, $param4=null) {

I'm looking to do:

public function index($params) {

  if (isset($params[0])) {

u_mulder · Accepted Answer · 2016-12-23 21:28:41Z

1

It is obviously:

call_user_func_array(
    [$this->controller, $this->method], 
    [$this->params]
);

answered Dec 23, 2016 at 21:28

u_mulder

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1 Comment

MultiDev Over a year ago

Obvious to you, not to me. I'm new to this. But thank you for your help!

Isa Ishangulyyev · Accepted Answer · 2016-12-23 21:33:04Z

0

Well I think you can get all of the arguments passed to a function via func_get_args. This just returns an array of all arguments without the need to specify expected args in the paranthesis.

answered Dec 23, 2016 at 21:33

Isa Ishangulyyev

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