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Home/ Questions/Q 4603872
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T00:12:52+00:00 2026-05-22T00:12:52+00:00

I have an array sorting function as follows: public function sortAscending($accounts) { function ascending($accountA,

  • 0

I have an array sorting function as follows:

public function sortAscending($accounts)
{
    function ascending($accountA, $accountB) {
        if ($accountA['AmountUntilNextTarget'] == $accountB['AmountUntilNextTarget']) {
            return 0;
        }
        return ($accountA['AmountUntilNextTarget'] < $accountB['AmountUntilNextTarget']) ? -1 : 1;
    }
    usort($accounts, $ascending);

    return $accounts;
}

Clearly this is not ideal as it is hard-coding the key to search for. I thought I would make this generic by passing the key as a param to outside function, however this is then out-of-scope in the inner function. I tried to get around this by using a closure, which would have access to the param, instead of an inner function as follows:

public function sortAscending($accounts, $key)
{
    $ascending = function($accountA, $accountB) {
        if ($accountsA[$key] == $accountB[$key]) {
            return 0;
        }
        return ($accountA[$key] < $accountB[$key]) ? -1 : 1;
    }
    usort($accounts, $ascending);

    return $accounts;
}

However usort() only accepts a function name, so this doesn’t work. Can anyone see a (better?) way of achieving this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-22T00:12:52+00:00Added an answer on May 22, 2026 at 12:12 am
    • When defining closures, you can use the use keyword to let the function “see” a certain variable (or variables). See also the PHP documentation about Anonymous functions.

    Closures may also inherit variables
    from the parent scope. Any such
    variables must be declared in the
    function header. Inheriting variables
    from the parent scope is not the same
    as using global variables. Global
    variables exist in the global scope,
    which is the same no matter what
    function is executing. The parent
    scope of a closure is the function in
    which the closure was declared (not
    necessarily the function it was called
    from).

    • Please also note that defining a closure and assigning it to a variable is a normal assignment operation, so you will need the ; after the closing } of the closure.

    After making these changes your code would look like this (and should work fine):

    public function sortAscending($accounts, $key)
    {
        $ascending = function($accountA, $accountB) use ($key) {
            if ($accountsA[$key] == $accountB[$key]) {
                return 0;
            }
            return ($accountA[$key] < $accountB[$key]) ? -1 : 1;
        };
        usort($accounts, $ascending);
    
        return $accounts;
    }
    
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