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, Sep 23 '14 at 11:54

score 175 · Accepted Answer · answered Sep 23 '14 at 11:49

175

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;
});

answered Sep 23 '14 at 11:49

Barmar

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1

**Note:** Requires PHP 5.3.x ... see also http://stackoverflow.com/a/25994339/42223 – dreftymac Feb 06 '17 at 22:05
1

@dreftymac 5.3 is required just for anonymous function syntax, it's not specific to the `use` modifier. – Barmar Feb 06 '17 at 22:09

score 13 · Answer 2 · answered Sep 23 '14 at 11:48

13

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 '14 at 11:48

php-dev

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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. – Barmar Feb 06 '17 at 22:11

score 9 · Answer 3 · answered Sep 23 '14 at 11:48

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 '14 at 11:48

Joe

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Use external variable in array_filter

3 Answers3