Laravel Tests pass to model to View

Question

I'm mocking my repository correctly, but in cases like show() it either returns null so the view ends up crashing the test because of calling property on null object.

I'm guessing I'm supposed to mock the eloquent model returned but I find 2 issues:

1. What's the point of implementing repository pattern if I'm gonna end up mocking eloquent model anyway

2. How do you mock them correctly? The code below gives me an error.

   $this->mockRepository->shouldReceive('find')
       ->once()
       ->with(1)
       ->andReturn(Mockery::mock('MyNamespace\MyModel)
   
           // The view may call $book->title, so I'm guessing I have to mock
           // that call and it's returned value, but this doesn't work as it says
           // 'Undefined property: Mockery\CompositeExpectation::$title'
           ->shouldReceive('getAttribute')
           ->andReturn('')
   );
   

Edit:

I'm trying to test the controller's actions as in:

$this->call('GET', 'books/1'); // will call Controller#show(1)

The thing is, at the end of the controller, it returns a view:

$book = Repo::find(1);
return view('books.show', compact('book'));

So, the the test case also runs view method and if no $book is mocked, it is null and crashes

Answer 1

So you're trying to unit test your controller to make sure that the right methods are called with the expected arguments. The controller-method fetches a model from the repo and passes it to the view. So we have to make sure that

• the find()-method is called on the repo
• the repo returns a model
• the returned model is passed to the view

But first things first:

What's the point of implementing repository pattern if I'm gonna end up mocking eloquent model anyway?

It has many purposes besides (testable) consisten data access rules through different sources, (testable) centralized cache strategies, etc. In this case, you're not testing the repository and you actually don't even care what's returned, you're just interested that certain methods are called. So in combination with the concept of dependency injection you now have a powerful tool: You can just switch the actual instance of the repo with the mock.

So let's say your controller looks like this:

class BookController extends Controller {

    protected $repo;

    public function __construct(MyNamespace\BookRepository $repo)
    {
        $this->repo = $repo;
    }

    public function show()
    {
        $book = $this->repo->find(1);

        return View::make('books.show', compact('book'));
    }
}

So now, within your test you just mock the repo and bind it to the container:

public function testShowBook()
{
    // no need to mock this, just make sure you pass something
    // to the view that is (or acts like) a book
    $book = new MyNamespace\Book;

    $bookRepoMock = Mockery::mock('MyNamespace\BookRepository');

    // make sure the repo is queried with 1
    // and you want it to return the book instanciated above
    $bookRepoMock->shouldReceive('find')
                 ->once()
                 ->with(1)
                 ->andReturn($book);

    // bind your mock to the container, so whenever an instance of
    // MyNamespace\BookRepository is needed (like in your controller),
    // the mock will be loaded.
    $this->app->instance('MyNamespace\BookRepository', $bookRepoMock);

    // now trigger the controller method
    $response = $this->call('GET', 'books/1');

    $this->assertEquals(200, $response->getStatusCode());

    // check if the controller passed what was returned from the repo
    // to the view
    $this->assertViewHas('book', $book);
}

//EDIT in response to the comment:

Now, in the first line of your testShowBook() you instantiate a new Book, which I am assuming is a subclass of Eloquent\Model. Wouldn't that invalidate the whole deal of inversion of control[...]? since if you change ORM, you'd still have to change Book so that it wouldn't be class of Model

Well... yes and no. Yes, I've instantiated the model-class in the test directly, but model in this context doesn't necessarily mean instance of Eloquent\Model but more like the model in model-view-controller. Eloquent is only the ORM and has a class named Model that you inherit from, but the model-class as itself is just an entity of the business logic. It could extend Eloquent, it could extend Doctrine, or it could extend nothing at all.

In the end it's just a class that holds the data that you pull e.g. from a database, from an architecture point of view it is not aware of any ORM, it just contains data. A Book might have an author attribute, maybe even a getAuthor() method, but it doesn't really make sense for a book to have a save() or find() method. But it does if you're using Eloquent. And it's ok, because it's convenient, and in small project there's nothing wrong with accessing it directly. But it's the repository's (or the controller's) job to deal with a specific ORM, not the model's. The actual model is sort of the outcome of an ORM-interaction.

So yes, it might be a little confusing that the model seems so tightly bound to the ORM in Laravel, but, again, it's very convenient and perfectly fine for most projects. In fact, you won't even notice it unless you're using it directly in your application code (e.g. Book::where(...)->get();) and then decide to switch from Eloquent to something like Doctrine - this would obviously break your application. But if this is all encapsulated behind a repository, the rest of your application won't even notice when you switch between databases or even ORMs.

So, you're working with repositories, so only the eloquent-implementation of the repository should actually be aware that Book also extends Eloquent\Model and that it can call a save() method on it. The point is that it doesn't (=shouldn't) matter if Book extends Model or not, it should still be instantiable anywhere in your application, because within your business logic it's just a Book, i.e. a Plain Old PHP Object with some attributes and methods describing a book and not the strategies how to find or persist the object. That's what repositories are for.

But yes, the absolute clean way is to have a BookInterface and then bind it to a specific implementation. So it could all look like this:

Interfaces:

---

interface BookInterface 
{
    /**
     * Get the ISBN.
     *
     * @return string
     */
    public function getISBN();
}

interface BookRepositoryInterface()
{
    /**
     * Find a book by the given Id.
     *
     * @return null|BookInterface
     */
    public function find($id);
}

Concrete implementations:

---

class Book extends Model implements BookInterface
{
    public function getISBN()
    {
        return $this->isbn;
    }
}

class EloquentBookRepository implements BookRepositoryInterface
{
    protected $book;

    public function __construct(Model $book)
    {
        $this->book = $book;
    }

    public function find($id)
    {
        return $this->book->find($id);
    }
}

And then bind the interfaces to the desired implementations:

App::bind('BookInterface', function() 
{
    return new Book;
});

App::bind('BookRepositoryInterface', function() 
{
    return new EloquentBookRepository(new Book);
});

It doesn't matter if Book extends Model or anything else, as long as it implements the BookInterface, it is a Book. That's why I bravely instantiated a new Book in the test. Because it doesn't matter if you change the ORM, it only matters if you have several implementations of the BookInterface, but that's not very likely (sensible?), I guess. But just to play it safe, now that it's bound to the IoC-Container, you can instantiate it like this in the test:

$book = $this->app->make('BookInterface');

which will return an instance of whatever implementation of Book you're currently using.

So, for better testability

• Code to interfaces rather than concrete classes
• Use Laravel's IoC-Container to bind interfaces to concrete implementations (including mocks)
• Use dependency injection

I hope that makes sense.