PHP: how to get array based on nested value

Question

Given this array and a name (e.g. 'def'), how do I get the containing array, or key?

$all = array(
 '0' => array(
      'name' => 'abc'
      'option' => 1,
    ),
 '1' => array(
      'name' => 'def'
      'option' => 1,
    ),
 '2' => array(
      'name' => 'ghi'
      'option' => 0,
    ),
);

What's the best way to return this array given 'def'?

$single = array(
  'name' => 'def'
  'option' => 1,
);

I could do something like this:

$single = array();
foreach ($all as $key => $value) {
  if ($value['name'] == 'def') {
    $single = $all[$key]; 
  }
}

Or prerender the keys in the array so that it looks like this:

$all = array(
 'abc' => array(
      'name' => 'abc'
      'option' => 1,
    ),
 'def' => array(
      'name' => 'def'
      'option' => 1,
    ),
 'ghi' => array(
      'name' => 'ghi'
      'option' => 0,
    ),
);
$single = $all['def'];

But I'm wondering if there's a shorter php function for that.

The foreach method was always the best for such cases (although it would be marginally better if you include a break inside the if block). You can do it in other ways as well, but they will be more verbose, or less clear, or both. — Jon
– Jon, Commented Jan 18, 2013 at 1:01

Matt Zeunert · Accepted Answer · 2013-01-18 01:03:56Z

4

You could use array_filter:

array_filter($array, function($var){
   return $var["name"] == "def";
});

answered Jan 18, 2013 at 1:03

Matt Zeunert

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2 Comments

Jon Over a year ago

But you also need to unwrap the value found, so something like list($single) = array_filter(...) since reset complains if you pass in a temporary (it takes its parameter by reference). That's the best I could come up with as well, but IMHO it's too clever for its own good.

Danny_Joris Over a year ago

Thanks for the nice ideas! I marked this as the answer because it's the closest to a single function call as possible. Though indeed not sure I'd use it because of readablity and bit complex.

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