返回JSON数据的接口解决方案

header('Content-type: application/json');
$type = isset($_GET['type']) ? intval($_GET['type']) : 1;

$array = array();
if ($type == 1) {
  $array = array("test" => "I'm 1");
} else if ($type == 2) {
  $array = array("test" => "I'm 2");
}

echo json_encode($array);