Using array_walk to alter array values

Question

Here's the function I'm trying to call on a set of array items:

function do_this_stuff( &$key ) {

    $lookups = array(
        'this' => 'that'
    );

    if (array_key_exists($key, $lookups)) {
        return $lookups[$key];
    } else {
        return ucwords(str_replace("_", " ", $key));
    }
}

And the call to it:

array_walk($data[$set][0], 'do_this_stuff');

If anything in the $lookups array is in the array in param one of array walk, I want to replace its contents. The do_this_stuff function works, but nothing I've tried has resulted in the actual input array values updating.

Downvoted. Hm. I feel like this is a reasonable question, easily overlooked in the docs. I can't be the only one. — Chords
– Chords, Commented Jan 29, 2014 at 17:58

Barmar · Accepted Answer · 2014-01-29 17:50:15Z

4

You need to assign the updated value back to $key, not return it.

function do_this_stuff( &$key ) {

    $lookups = array(
        'this' => 'that'
    );

    if (array_key_exists($key, $lookups)) {
        $key = $lookups[$key];
    } else {
        $key = ucwords(str_replace("_", " ", $key));
    }
}

answered Jan 29, 2014 at 17:50

Barmar

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5 Comments

Chords Over a year ago

Thanks, Barmar! That did the trick. Kind of renders this function useless though, no? I'm new to these functional paradigms.

Barmar Over a year ago

I'm not sure why you say that. If you mean that you could do the same thing with a for or foreach loop, that's true. Sometimes there's more than one way to skin a cat.

Chords Over a year ago

If I wanted to use do_this_stuff outside of array_walk, I would want the function to return or echo, rather than just set equal to $key. Seems like something that just hasn't clicked for me.

Barmar Over a year ago

But the reason it takes the argument as a reference is so that it can modify what it was given, rather than return or echo it. You could split it into two functions: one that returns the replacement, and another that calls the first function and modifies its argument with what it returns.

Chords Over a year ago

Ah, that makes sense. Thank you very much for sticking around and explaining!

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