How to implement array_filter() with a dynamic/passed-in filtering value

Question

I have this little function to filter my array by keys:

 private function filterMyArray( )
 {
      function check( $v )
      {
           return $v['type'] == 'video';
      }
      return array_filter( $array, 'check' );
 }

This works great but since I have more keys to filter, I was thinking in a way to pass a variable from the main function: filterMyArray($key_to_serch) without success, also I've tried a global variable, but seems not work.

Due some confusion in my question :), I need something like this:

 private function filterMyArray( $key_to_serch )
 {
      function check( $v )
      {
           return $v['type'] == $key_to_serch;
      }
      return array_filter( $array, 'check' );
 }

Any idea to pass that variable?

mmm ... nested functions ... not pretty, especially when php provides functions that do this for you. try array_map() or array_filter() with lambda/closure — user895378
– user895378, Commented Dec 9, 2011 at 1:40

Olivier Pons · Accepted Answer · 2014-04-20 07:42:43Z

3

This is where anonymous functions in PHP 5.3 come in handy (note the use of use):

private function filterMyArray($key)
{
     return array_filter(
         $array,
         function check($v) use($key) {
             return $v['type'] == $key;
         }
     );
}

edited Apr 20, 2014 at 7:42

Olivier Pons

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answered Dec 9, 2011 at 1:39

user142162

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1 Comment

Olivier Pons Over a year ago

This is where identation comes in place: this code is very hard to understand (for me) if it's not properly indented. I've just modified identation, roll back if you think yours is easier to read

hackartist · Accepted Answer · 2011-12-09 01:41:12Z

0

private function filterMyArray($key_to_search) {
  function check( $v ) {
       return $v[$key_to_search] == 'video';
  }
  return array_filter( $array, 'check' );
}

should work because the inner function has access to the variables in the outer function

answered Dec 9, 2011 at 1:41

hackartist

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Comments

middus · Accepted Answer · 2011-12-09 01:41:47Z

0

You need to use the use keyword to get the variable in scope, c.f. this example in php's doc.

answered Dec 9, 2011 at 1:41

middus

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Comments

nickb · Accepted Answer · 2011-12-09 01:43:08Z

-1

Here's a PHP < 5.3 version using create_function.

private function filterMyArray( $key)
{
      $fn = create_function( '$v', "return $v[$key] == 'video';");
      return array_filter( $array, $fn);
}

answered Dec 9, 2011 at 1:43

nickb

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1 Comment

Jon Over a year ago

Actually, this will not work because the dollar signs inside the double-quoted string will be seen as variable references.

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