nico nico - 3 months ago 16
PHP Question

array_filter and multidimensional array

Say I have an array such as:

$arr[] = array("id" => 11, "name" => "First");
$arr[] = array("id" => 52, "name" => "Second");
$arr[] = array("id" => 6, "name" => "Third");
$arr[] = array("id" => 43, "name" => "Fourth");


I would like to get the name correspondent to a certain ID so that I can do:

$name = findNameFromID(43);


and get, for instance, "Fourth".

I thought of using
array_filter
but I am a bit stuck on how to correctly pass a variable. I have seen questions such as this one but I don't seem to be able to extend the solution to a multidimensional array.

Any help?

Answer
findNameFromID($array,$ID) {
     return array_values(array_filter($array, function($arrayValue) use($ID) { return $arrayValue['id'] == $ID; } ));
}

$name = findNameFromID($arr,43);
if (count($name) > 0) {
    $name = $name[0]['name'];
} else {
    echo 'No match found';
}

PHP 5.3.0 and above

EDIT

or variant:

findNameFromID($array,$ID) {
    $results = array_values(array_filter($array, function($arrayValue) use($ID) { return $arrayValue['id'] == $ID; } ));
    if (count($results) > 0) {
        return $name[0]['name'];
    } else {
        return FALSE;
    }
}

$name = findNameFromID($arr,43);
if (!$name) {
    echo 'No match found';
}

EDIT #2

And from PHP 5.5, we can use array_column()

findNameFromID($array, $ID) {
    $results = array_column($array, 'name', 'id');
    return (isset($results[$ID])) ? $results[$ID] : FALSE;
}
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