Undefined variable inside PHP array_filter [duplicate]

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
– ElefantPhace, Commented Sep 1, 2015 at 4:50
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
– Rasclatt, Commented Sep 1, 2015 at 4:50
try this $response['response'] = array_filter($response['response'], function($tier) use $leagueKey { return ($tier['LeagueKey'] === $leagueKey ? true : false); }); — Ganesh Ghalame
– Ganesh Ghalame, Commented Sep 1, 2015 at 4:52
Shouldnt be marked as duplicate.... This is a good question and different than the other one. Just a similar answer. — Andrew
– Andrew, Commented Nov 5, 2015 at 14:07

Scopey · Accepted Answer · 2020-02-07 19:53:15Z

37

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
);

edited Feb 7, 2020 at 19:53

answered Sep 1, 2015 at 4:53

Scopey

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Comments

Ganesh Ghalame · Accepted Answer · 2015-09-01 05:05:05Z

5

try this

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

edited Sep 1, 2015 at 5:05

answered Sep 1, 2015 at 4:54

Ganesh Ghalame

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2 Comments

Sverri M. Olsen Over a year ago

You are missing parens, use ($leagueKey).

Ganesh Ghalame Over a year ago

@Sverri M. Olsen thanks, it won't work without brackets

Community · Accepted Answer · 2017-05-23 10:27:45Z

2

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

edited May 23, 2017 at 10:27

CommunityBot

11 silver badge

answered Sep 1, 2015 at 4:52

DannyPhantom

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Comments

Ath Piseth · Accepted Answer · 2015-09-01 05:13:27Z

-4

try globals vairable:

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

answered Sep 1, 2015 at 5:13

Ath Piseth

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1 Comment

Ganesh Ghalame Over a year ago

Using global is not a good practice check this for clarification stackoverflow.com/questions/1557787/…

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