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Home/ Questions/Q 8000445
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Editorial Team
  • 0
Editorial Team
Asked: June 4, 20262026-06-04T15:47:48+00:00 2026-06-04T15:47:48+00:00

If I have this array, ini_set(‘display_errors’, true); error_reporting(E_ALL); $arr = array( ‘id’ => 1234,

  • 0

If I have this array,

ini_set('display_errors', true);
error_reporting(E_ALL);

$arr = array(
  'id' => 1234,
  'name' => 'Jack',
  'email' => 'jack@example.com',
  'city' => array(
    'id' => 55,
    'name' => 'Los Angeles',
    'country' => array(
      'id' => 77,
      'name' => 'USA',
     ),
  ),
);

I can get the country name with

$name = $arr['city']['country']['name'];

But if the country array doesn’t exist, PHP will generate warning:

Notice: Undefined index ... on line xxx

Sure I can do the test first:

if (isset($arr['city']['country']['name'])) {
  $name = $arr['city']['country']['name'];
} else {
  $name = '';  // or set to default value;
}

But that is inefficient. What is the best way to get $arr['city']['country']['name']
without generating PHP Notice if it doesn’t exist?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-04T15:47:50+00:00Added an answer on June 4, 2026 at 3:47 pm

    I borrowed the code below from Kohana. It will return the element of multidimensional array or NULL (or any default value chosen) if the key doesn’t exist.

    function _arr($arr, $path, $default = NULL) 
    {
      if (!is_array($arr))
        return $default;
    
      $cursor = $arr;
      $keys = explode('.', $path);
    
      foreach ($keys as $key) {
        if (isset($cursor[$key])) {
          $cursor = $cursor[$key];
        } else {
          return $default;
        }
      }
    
      return $cursor;
    }
    

    Given the input array above, access its elements with:

    echo _arr($arr, 'id');                    // 1234
    echo _arr($arr, 'city.country.name');     // USA
    echo _arr($arr, 'city.name');             // Los Angeles
    echo _arr($arr, 'city.zip', 'not set');   // not set
    
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