CODEWITHSUNDEEP
phpauthenticationlaravelmodeleloquent
Peter
Peter

Reputation: 2711

how display for unlogged user output of a relation which haswhere clasue with logged user ID

I have this relation, which contains reference to logged user

public function booksdislikedbyUser()
{
return $this->hasMany('Userattitude')->where('creator_id', Auth::user()->id)->selectRaw('count(IF(attitude = -1,0,1)) as dislikedbooks')->groupBy('entity_id');

}

My VIEW file which displays the output must be visible both for logged and unlogged user.

Problem: if user is not logged, I receive this error: Trying to get property of non-object

My question: what is best practice if I want to get the view and disable(?) the relation?

Option 1: What I am about to do is to put two variants of the code in IF - ELSE brackets.

if(Auth::check())

It works, but it doesn.t look like a good practice.

Option 2: in place of 

->where('creator_id', Auth::user())

i would put 

->where('creator_id', ($user_id ? $user_id : NULL))

Where I shoud define the variable then? $user_id = Auth::user()->id;

Upvotes: 0

Views: 87

Answers (1)

lukasgeiter
lukasgeiter

Reputation: 153070

I'd go with Option 2.

public function booksdislikedbyUser()
{
    $userId = (Auth::user() ? Auth::user()->id : null);
    return $this->hasMany('Userattitude')->where('creator_id', $userId)->selectRaw('count(IF(attitude = -1,0,1)) as dislikedbooks')->groupBy('entity_id');
}

Or even simpler, use the Auth::id() function:

return $this->hasMany('Userattitude')->where('creator_id', Auth::id())->selectRaw('count(IF(attitude = -1,0,1)) as dislikedbooks')->groupBy('entity_id');

Upvotes: 1

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