你会怎么做这个PHP代码？

Question

Is there a better way to write the below, other then using a switch or if/else statement? Like is this a situation where a PHP's varaiable variable ( $$var ) could come to use? If so how would you write this code?

$type = 2;

switch ($type) {
 case 1:
   $type = 'gif';
   break;
 case 2:
   $type = 'jpg';
   break;
 case 3:
   $type = 'png';
   break;
 default:
   $type = 'jpg';
   break;
}

Answer 1

$types = array(1 => 'gif', 2 => 'jpg', 3 => 'png', 4 => 'jpg');

...

array_key_exists($type, $types) ? $types[$type] : 'jpg';