PHP - array_filter with reference Or similar to it

Question

I have an array like below.

$a = array(
            array("id" => 1, "name" => "njx", "count" => 0),
            array("id" => 2, "name" => "peter", "count" => 4),
            array("id" => 3, "name" => "jeffry", "count" => 2),
            array("id" => 4, "name" => "adam", "count" => 6)
          );

and applied filter like this.

$fa = array_filter($a, function ($item) {
    return ($item["count"] > 0);
});

then i applied usort on variable $fa. After that i loop through $fa and assigned some values, but they are not get reflected in variable $a.

something like below,

usort($fa, 'comp');

foreach ($fa as &$t) {
    $t["score"] = 120;
}

var_dump($a); //this doesn't contain "score" field.

So my questions is how to get filtered array with original array reference?

It would be more easy for others to understand if you show a snippet of output you want. — Bhavik Shah
– Bhavik Shah, Commented Nov 13, 2013 at 10:48
As long as array_filter will preserve keys, you could do foreach($fa as $faKey=>$dummy) $a[$faKey]['score']=120; — sofl
– sofl, Commented Nov 13, 2013 at 10:57
So it needs another loop over $fa and assign score based on id check? Can't i get filtered array that still hold reference to $a? — Sathya
– Sathya, Commented Nov 13, 2013 at 11:01

jari · Accepted Answer · 2013-11-13 11:18:46Z

4

array_filter returns a new array and not a reference, thats why any changes applied to $fa wont be reflected in $a.

Instead of using array_filter you could use a foreach loop like this:

foreach($a as &$t) {
    if($t['count'] > 0) {
        $t['score'] = 120;
    }
}

And then sort using usort.

edited Nov 13, 2013 at 11:18

answered Nov 13, 2013 at 11:09

jari

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