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array filter – Undefined variable inside PHP array_filter

Posted by: admin July 12, 2020 Leave a comment

Questions:

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!

How to&Answers:

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
);

Answer:

try this

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

Answer:

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

Answer:

try globals vairable:

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