Question

逻辑：用户有很多信用。

控制器：

public function getAdminAccount()
{
    $company = User::find(Auth::user()->id)
                ->companies()
                ->with('users.credits')
                ->first();
    $this->layout->content = View::make('agents.admin.account',
                array('company' => $company));
}

我的控制器功能正在返回我需要的JSON数据中的正确数据，但是，我希望按照user_id计算结果，最好在它们在视图中显示之前。

JSON：

[
   {
      "id":"4",
      "user_id":"4",
      "credits":"10",
      "created_at":"2013-09-30 09:39:25",
      "updated_at":"0000-00-00 00:00:00"
   },
   {
      "id":"67",
      "user_id":"4",
      "credits":"6",
      "created_at":"2013-08-05 12:40:25",
      "updated_at":"0000-00-00 00:00:00"
   }
][
   {
      "id":"6",
      "user_id":"6",
      "credits":"9",
      "created_at":"2013-06-25 16:24:18",
      "updated_at":"0000-00-00 00:00:00"
   },
   {
      "id":"69",
      "user_id":"6",
      "credits":"2",
      "created_at":"2014-01-16 10:09:55",
      "updated_at":"0000-00-00 00:00:00"
   }
]]

所以，对于那个JSON结果，应该说：

user_id：4，总积分：16
user_id：6，总积分：11

据我所知，在控制器中格式化JSON是有益的，而不是视图。我尝试了->sum()，->groupBy()功能，但没有成功。

谢谢。

更新

public function getAdminAccount()
{
    $company = User::find(Auth::user()->id)
                ->companies()
                ->with('users.credits')
                ->get()
                ->sum('credits');
    $this->layout->content = View::make('agents.admin.account', array('company' => $company));
}

以上功能将积分总数加在一起，非常接近我所需要的！我只需要构成总数的每个用户的积分数量！

更新

使用此forloop似乎返回正确的MYSQL逻辑，但仍然没有正确的值：

@foreach ($company->users as $user)
{{ $user->credits()->count('credits') }}<br>
@endforeach

MYSQL：

select count(`credits`) as aggregate from `usages` where `usages`.`user_id` = '4'

Answer 1

您需要的是sum而非count，但无论如何在循环中使用它意味着n + 1问题，因此您最好这样做：

// User model
public function creditsSum()
{
    return $this->hasOne('Credit')
        ->selectRaw('user_id, sum(credits) as sum') // user_id must be in select to match the models
        ->addSelect(DB::raw('count(*) as count')) // that just in case you also need 
                                                  // credits count, if not remove
        ->groupBy('category_id');
}

// Then you can access it as easily as every relation:
User::with('creditsSum')->get();

// you will have eager loaded data on each user model:

$user->credits_sum->sum;
$user->credits_sum->count; // if you select it too

Answer 2

如果我以正确的方式预测您的数据库布局，我认为连接是最快的解决方案。这样您就不需要为每个用户触发查询。它也是SUM而不是COUNT，count只计算数据集的数量。

$company->users()->join('credits', 'users.id', '=', 'credits.user_id')
                 ->selectRaw('SUM(credits.credits) as `total_credits`, users.username')
                 ->groupBy('credits.user_id')->get();

这应该给你类似

的东西

total_credits | username
------------------------
           24 | John
           12 | Michael

如果您还希望获得没有任何积分的用户，则应使用leftJoin。

Laravel计数导致控制器或视图

2 个答案: