如何从视图访问存储库逻辑？

Question

Basically, I'm writing a simple blogging application where users can vote up or down on posts (I've named this process scoring inside my application). My problem is I am not sure what the best approach is for accessing a method to determine if the user has voted or not on the post, as I don't want to pass a repository into my view nor do I want the model to have methods mimicking the repository... Here are those two ideas - are these the only approaches?

The first approach requires that I pass a PostRepository to my views as well as the Post model which is already passed...

<!-- Repository-in-view approach -->
<p>You voted {{ $postRepository->hasUserScored($post->id, $user->id) ? 'up' : 'down' }}.</p>

-----------

// Inside `PostRepository`
public function hasUserScored($postId, $userId, $vote = true)
{
    // DB logic to determine ...
}

Or should I maybe do something like this?

<!-- Repository-in-view approach -->
<p>You voted {{ $post->hasUserScored($user->id) ? 'up' : 'down' }}.</p>

-----------

// Inside `Post`
public function hasUserScored($userId)
{
    return (new PostRepository)->hasUserScored($this->id, $userId);
}

// Inside `PostRepository`
public function hasUserScored($postId, $userId, $vote = true)
{
    // DB logic to determine ...
}

What is the best way to overcome this? Any help greatly appreciated, thanks!

Answer 1

Best practice is to avoid calling functions from within views.

Either call the function from the controller, and pass that data to the view. Or, alternatively, use a view composer.

Answer 2

Why do you do this in such a complicated way? Can't your Post model just have many-to-many relationship with users + pivot table? Then, checking a User vote is as simple as:

function users()
{
    return $this->belongsToMany('User')->withPivot('vote');       
}

function getUserScore($userId)
{
     return $this->users->find($userId)->pivot->vote;       
}

Also, if you want to check whether a user has voted or not, simply do:

function hasUserVoted($userId)
{
     return $this->users->contains($userId);
}

Answer 3

you have to pass the Repository to the view.. although what you want to do is not the smartest approach but according with what you need.. the way is

on your controller __construct you add something like this

function __construct(PostRepositoryInterface $post){
    $this->post = $post;
}

if you dont use interface then just pass the repository

then you return to the view

view(your.view)->with('post', $this->post);

then you are done..