Nested function with PHP

Question

I'm writing customized sort function. Current (non-working) code looks like this:

<?php
 function sort_by_key($array, $key) {

   function custom_compare ($a, $b) {
     if ($a[$key][0] > $b[$key][0]) { return 1; }
     else { return -1; }
     }
  return usort($array, "custom_compare");
 }
?>

The problem is that I cannot pass $key variable to custom_compare function. I'd like to avoid using global variables (ugly coding).

With further reading I noticed that PHP doesn't support nested functions. My code returns error when run second time, PHP tries to re-declarate custom_compare -function. Is there any other options than global variables. I cannot pass more arguments to custom_compare because it is used by usort. — ex4
– ex4, Commented Sep 17, 2016 at 10:48

Progrock · Accepted Answer · 2016-09-17 11:28:13Z

5

Untested, but you could use an anonymous function:

<?php
 function sort_by_key($array, $key) {

   $custom_compare = function ($a, $b) use ($key) {
     if ($a[$key][0] > $b[$key][0]) { return 1; }
     else { return -1; }
     };

  return usort($array, $custom_compare);
 }

Based on a small modification of your existing function.

Further your function still needs a small change:

<?php
function sort_by_key(&$array, $key) {
    $custom_compare = function ($a, $b) use ($key) {
        if ($a[$key][0] > $b[$key][0]) {
            return 1;
        } else {
            return -1;
        }
    };

    usort($array, $custom_compare);
}

$array = array(
    array(
        'foo' => array(
            2
        )
    ),
    array(
        'foo' => array(
            3
        )
    ),
    array(
        'foo' => array(
            1
        )
    )
);

sort_by_key($array, 'foo');
var_export($array);

Output:

array (
  0 => 
  array (
    'foo' => 
    array (
      0 => 1,
    ),
  ),
  1 => 
  array (
    'foo' => 
    array (
      0 => 2,
    ),
  ),
  2 => 
  array (
    'foo' => 
    array (
      0 => 3,
    ),
  ),
)

edited Sep 17, 2016 at 11:28

answered Sep 17, 2016 at 10:45

Progrock

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2 Comments

ex4 Over a year ago

Thanks for reply, but that doesn't work. Gives following error instead: PHP Parse error: syntax error, unexpected 'use' (T_USE), expecting '{'

ex4 Over a year ago

Sorry.... My bad... A stupid error, I did function custom_compare($a, $b) use ($key), not $custom_compare = function($a, $b) use ($key)

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