PHP variable scope issue inside of array_filter

Question

I come from a JS world and have trouble understanding this piece of code:

function getCountryName($code, $codes){
  print_r($code);

  array_filter(
    $codes,
    function ($e) {
      if($e->cca2 == $code){
        print_r($e->name->common);
      }
    }
  );
}

The second line is for debug only and it prints given $code. But here: if($e->cca2 == $code){ $code seems to be undefined. How is it even possible? It works with $GLOBALS['code'] but I don't want to use globals. How come array_filter has no access to it's own scope?

ishegg · Accepted Answer · 2017-10-22 13:56:32Z

3

$code is in getCountryName()’s scope. You then have a callback for array_filter(), which is an anonymous function and has its own scope. To access the variable, you need to use it, which will make the callback inherit the variable from its parent scope (the function):

function getCountryName($code, $codes){
print_r($code);
 array_filter(
    $codes,
    function ($e) use ($code) {
      if($e->cca2 == $code){
        print_r($e->name->common);
      }
    }
  );
}

answered Oct 22, 2017 at 13:56

ishegg

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Comments

Hemen Ashodia · Accepted Answer · 2017-10-22 13:56:04Z

1

you need to use 'use'.

function getCountryName($code, $codes){
  print_r($code);

  array_filter(
    $codes,
    function ($e) use($code) {
      if($e->cca2 == $code){
        print_r($e->name->common);
      }
    }
  );
}

answered Oct 22, 2017 at 13:56

Hemen Ashodia

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Comments

MatsLindh · Accepted Answer · 2017-10-22 13:56:44Z

1

Because function() creates its own scope. If you want to pass a variable from the outside into the inner scope, you can use use:

array_filter(
  $codes,
  function ($e) use ($code) {
    if($e->cca2 == $code){
      print_r($e->name->common);
    }
  }
);

This way the use of the variable from the outside scope is explicit.

answered Oct 22, 2017 at 13:56

MatsLindh

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