Laravel 5.4-表单/ API /视图

Question

I created an API ( store ) that saves the data on the database and returns 201 if successful or 404 if not.

if ($visit->save()){
    $visit->view_visit = [
    'href' => 'api/v1/visit/' . $visit->id,
    'method' => 'GET'
     ];
     $response = [
         'msg' => 'Visit created.',
         'visit' => $visit
     ];
     return response()->json($response, 201);
}
$response = [
    'msg' => 'Error during creation.'
    ];
return response()->json($response, 404);

It works perfectly. Using postman you can see that the status will be <<201 Created>>.

This API should be used in two ways: called by another application or called by a Laravel form. This is the question:

How do I call it in a way if it successful, it will load a given view on the browsers? In other words, is there a way to make the form call a route (the api itself, something like ../api/visit/) and in case of success loads the other view? Also, I would like to pass the content of response['msg'] to this new view.

I know I could do it inside the store method by filtering the HTTP referrer, but I would like to keep the controller code strictly to manage the record creation. Besides that, I have to send the 201/404 codes along with the returned data.

I also considered creating another controller to handle the API response and then call the form, but it still sounds too much -- it's supposed to be easy, I guess.

Answer 1

In laravel you can use a helpful method which determines if the request that has been sent is an AJAX request or just a normal request, which is:

$request->wantsJson()

So, Inside your controller in the return function, you will make an if statement:

if ($request->wantsJson()) {
  return response()->json();
}else{
  return view(...);
}