pass argument to filter function in PHP

Question

at one point in my function I need to filter an array based on a last digit of a key, but I cannot pass argument to the function I'm using to filter

here's the code:

    function functionName($number){
        //some code
        $items = array_filter(SomeClass::$items, function ($item, $number){
            return ($item%10 == $number);
        }, ARRAY_FILTER_USE_KEY);
        //some code
    }

what can I do to make sure the inner function uses $number passed to functionName()?

@Carcigenicate edited, maybe OP was thinking about closures – nha May 16 '18 at 11:54 — nha, May 16 '18 at 11:54
function ($item, $number) use ($yourvar) {... – Eakethet May 16 '18 at 11:54 — Eakethet, May 16 '18 at 11:54

score 0 · Answer 1 · answered May 16 '18 at 11:57

0

You can pass variables to closure function by "use" Eg.

function ($var) use ($yourVarThatYouWannaPass) { return $var;}

answered May 16 '18 at 11:57

Eakethet

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pass argument to filter function in PHP

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