Undefined variable inside PHP array_filter

Question

this could be a very silly question but I just can't understand how PHP scope is working on this piece of code:

$leagueKey = 'NFL';
$response['response'] = array_filter($response['response'], function($tier){
    return ($tier['LeagueKey'] === $leagueKey ? true : false);
});

When I run that, I get an "Undefined variable: leagueKey" exception. On the other hand, this works perfectly well:

$response['response'] = array_filter($response['response'], function($tier){
    return ($tier['LeagueKey'] === 'NFL' ? true : false);
});

Why can't PHP see my $leagueKey variable inside the array_filter function?

Thanks!

because `$leagueKey` is defined outside of that function. In order to use it you could use `global $leagueKey` inside your anonymous function but it's not the best of ways — ElefantPhace, Sep 01 '15 at 04:50
Because it's out of scope, just like any function, you would have to feed an outside argument into it. `$leagueKey` is outside of the function, the string `'NFL'` is in, so it works. — Rasclatt, Sep 01 '15 at 04:50
try this $response['response'] = array_filter($response['response'], function($tier) *use $leagueKey* { return ($tier['LeagueKey'] === $leagueKey ? true : false); }); — Ganesh Ghalame, Sep 01 '15 at 04:52
Shouldnt be marked as duplicate.... This is a good question and different than the other one. Just a similar answer. — Andrew, Nov 05 '15 at 14:07

Scopey · Accepted Answer · 2020-02-07T19:53:15.570

Your $leagueKey variable is outside the scope of the anonymous function (closure). Luckily, PHP provides a very simple way to bring variables into scope - the use keyword. Try:

$leagueKey = 'NFL';
$response['response'] = array_filter($response['response'], function($tier) use ($leagueKey) {
    return $tier['LeagueKey'] === $leagueKey;
});

This is just telling your anonymous function to "use" the $leagueKey variable from the current scope.

Edit

Exciting PHP 7.4 update - we can now use "short closures" to write functions that don't have their own scope. The example can now be written like this (in PHP 7.4):

$response['response'] = array_filter(
    $response['response'], 
    fn($tier) => $tier['LeagueKey'] === $leagueKey
);

Ganesh Ghalame · Answer 2 · 2015-09-01T05:05:05.083

5

try this

$response['response'] = array_filter($response['response'], function($tier) use ($leagueKey) {
 return ($tier['LeagueKey'] === $leagueKey ? true : false); 
});

edited Sep 01 '15 at 05:05

answered Sep 01 '15 at 04:54

Ganesh Ghalame

3,817
3
21
27

1

You are missing parens, `use ($leagueKey)`. – Sverri M. Olsen Sep 01 '15 at 05:03
@Sverri M. Olsen thanks, it won't work without brackets – Ganesh Ghalame Sep 01 '15 at 05:08

score 2 · Answer 3 · edited May 23 '17 at 10:27

This is how the scope of all the variables work. Your anonymous function doesn't know anything about variables outside of it. This holds for all kind of functions: if you need to use a variable that is outside of a function - you pass it to the function. In this case you can't pass it anything, but if you are running PHP5.3+ you can do function($tier) use ($leagueKey){ which will tell the function that it needs to use $leagueKey defined outside of it. If your php is lower than 5.3 you'll have to use workaround like this one: link

score -4 · Answer 4 · answered Sep 01 '15 at 05:13

-4

try globals vairable:

$GLOBALS['leagueKey'] = 'NFL';
$response['response'] = array_filter($response['response'], function($tier){
    return ($tier['LeagueKey'] === $leagueKey ? true : false);
});

answered Sep 01 '15 at 05:13

Ath Piseth

75
7

Using global is not a good practice check this for clarification http://stackoverflow.com/questions/1557787/are-global-variables-in-php-considered-bad-practice-if-so-why – Ganesh Ghalame Sep 01 '15 at 05:25

Undefined variable inside PHP array_filter

4 Answers4