Use external variable in array_filter

Question

I've got an array, which I want to filter by an external variable. The situation is as follows:

$id = '1';
var_dump($id);
$foo = array_filter($bar, function($obj){
    if (isset($obj->foo)) {
        var_dump($id);
        if ($obj->foo == $id) return true;
    }
    return false;
});

The first var_dump returns the ID (which is dynamically set ofcourse), however, the second var_dump returns NULL.

Can anyone tell me why, and how to solve it?

@php-dev We can say it's definitely an array, because it's not throwing a PHP error. It may not always definitely be an array, depending on the code, but it's definitely an array in the use-case where he gets the error described in the OP — Joe
– Joe, Commented Sep 23, 2014 at 11:54

theschitz · Accepted Answer · 2023-09-27 14:53:42Z

189

The variable $id isn't in the scope of the function. You need to use the use clause to make external variables accessible:

$foo = array_filter($bar, function($obj) use ($id) {
    if (isset($obj->foo)) {
        var_dump($id);
        if ($obj->foo == $id) return true;
    }
    return false;
});

See documentation for Anonymous functions (Example #3 "Inheriting variables from the parent scope").

edited Sep 27, 2023 at 14:53

theschitz

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answered Sep 23, 2014 at 11:49

Barmar

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2 Comments

dreftymac Over a year ago

Note: Requires PHP 5.3.x ... see also stackoverflow.com/a/25994339/42223

Barmar Over a year ago

@dreftymac 5.3 is required just for anonymous function syntax, it's not specific to the use modifier.

php-dev · Accepted Answer · 2014-09-23 11:48:40Z

12

Variable scope issue!

Simple fix would be :

$id = '1';
var_dump($id);
$foo = array_filter($bar, function($obj){
    global $id;
    if (isset($obj->foo)) {
        var_dump($id);
        if ($obj->foo == $id) return true;
    }
    return false;
});

or, since PHP 5.3

$id = '1';
var_dump($id);
$foo = array_filter($bar, function($obj) use ($id) {
    if (isset($obj->foo)) {
        var_dump($id);
        if ($obj->foo == $id) return true;
    }
    return false;
});

Hope it helps

answered Sep 23, 2014 at 11:48

php-dev

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1 Comment

Barmar Over a year ago

I don't think the first version would work pre-5.3, since anonymous functions were added in 5.3. Earlier, you'd have to use the create_function() function, along with the global declaration.

Joe · Accepted Answer · 2014-09-23 11:48:25Z

9

Because your closure function can't see $id. You need the use keyword:

$foo = array_filter($bar, function($obj) use ($id) {

answered Sep 23, 2014 at 11:48

Joe

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