Ranking a user based on referrals

    User::orderBy('referrals')->get();

Could it be as easy as this?

@dfaux: I want to get a ranking for a specific user. Is that possible?

Ok I am going to give this a shot using query builder, don't have access to a test environment so this is a stab in the dark :)

    $rank = DB::table('users')->select(DB::raw('count(*) as rank'))
        ->where('referrals', '>', function($query) use ($user)
        {
            $query->select('referrals')->from('users')->where('id', $user->id)
        })->first();

If this does work you might need to increment the rank, test if this is the case.

@dfaux: Hm, doesn't seem to return the correct rank for a user. Just 0/1.

So using the referral count on your users model, you can get the rank of the current user like this:

    public function getRankAttribute()
    {
        $count = \DB::table('users')->where('referrals', '>', $this->referrals)->count();
        return $count ? $count + 1 : NULL;
    }

With this on your model, you can get the rank with $user->rank

@Woodlandtrek That really isn’t optimal, that’s going to execute an additional query for every user that comes back as part of a collection, or in simple terms: the N+1 problem.

@martinbean yeah, I understand it's not good for large groups. @cml123 shared the context on IRC and was only going to be using it for individual records.

An accessor like that won't be called just because the parent object is returned in a collection, will it? Only if the method is called explicitly?