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 $arr['city']['country']['name']
without generating PHP Notice if it doesn't exist?
Answer
I borrowed the code below from Kohana. It will return the element of multidimensional array or NULL (or any default value chosen) if the key doesn't exist.
function _arr($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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