Using anonymous functions to generate arrays in PHP

Question

I have this code chunk

        $data['cp'][$key]->prominence           = $meta_data['prominence'];
        $data['cp'][$key]->related_link = (function()
        {   $arr = array();
            for ( $i = 1; $i < 4 ; $i++ ) {
                $rldata = array();
                $rldata['title'] = $metadata['related_link_'.$i.'_title'];
                $rldata['title'] = $metadata['related_link_'.$i.'_url'];
                array_push( $arr, $rldata );
                }
            return  $arr;
        });

As you can see, I want $data['cp'][$key]->related_link to be equal to a multidimensional array dynamically generated by the anonymous function.

However when using print_r it just shows the key as being equal to a Closure. How do I edit the code so it actually returns the array, rather than just being equal to a closure.

On first blush, put a () at the end of your function definition: (function() { ... whatever ... })(); Nope, javascript on the brain. Use call_user_func. Example coming. — bishop
– bishop, Commented Nov 6, 2013 at 18:59

bishop · Accepted Answer · 2013-11-06 19:02:11Z

3

How about this:

<?php

$x = (function () { return array (1, 2, 3); });
$y = call_user_func(function () { return array (1, 2, 3); });
print_r($x);
print_r($y);

?>

results:

Closure Object
(
)
Array
(
    [0] => 1
    [1] => 2
    [2] => 3
)

answered Nov 6, 2013 at 19:02

bishop

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Comments

Miguel G. Flores · Accepted Answer · 2013-11-06 19:06:44Z

1

You should execute the function, until now you are just declaring it

$data['cp'][$key]->prominence   = $meta_data['prominence'];
$data['cp'][$key]->related_link = call_user_func(function()
            {   $arr = array();
                for ( $i = 1; $i < 4 ; $i++ ) {
                    $rldata = array();
                    $rldata['title'] = $metadata['related_link_'.$i.'_title'];
                    $rldata['title'] = $metadata['related_link_'.$i.'_url'];
                    array_push( $arr, $rldata );
                    }
                return  $arr;
            });

answered Nov 6, 2013 at 19:06

Miguel G. Flores

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