Laravel 5.4 pluck

Question

我在为广播新闻做作业。试图返回一个特定新闻节目的所有作业的视图，我就陷入了困境。

表

用户表

 id | name
 ---|-------
  1 | Admin
  2 | Susan
  3 | Ed
  4 | Jen

新闻广播台

 id | forStation_id | name
 ---|---------------|-----
  1 |       1       | AM
  2 |       1       | PM
  3 |       2       | Sports

台站表

 id | calls
 ---| -----
  1 | JNDV
  2 | YXWQ

赋值表

 id | anchorId | newscastId |  startDate  |   endDate   | isTemp
 ---|----------|------------|-------------|-------------|--------
  1 |     2    |     1      | 01 May 2017 | 31 Dec 2999 |
  2 |     3    |     1      | 02 May 2017 | 06 May 2017 |  True
  3 |     4    |     2      | 01 Apr 2017 | 31 Dec 2999 |
  4 |     3    |     3      | 01 Apr 2017 | 28 Apr 2017 |

(部分)分配模型

public function anchor()
{
    return $this->belongsTo(User::class, 'anchor_id')->withTrashed();
}

public function cast()
{
    return $this->belongsTo(Newscast::class, 'cast_id')->withTrashed();
}

(部分) Newscast模型

public function for_station()
{
    return $this->belongsTo(Station::class, 'for_station_id')->withTrashed();
}

function getNameInputStationAttribute() {
    return $this->for_station->calls . "-" . $this->name_input;
}

(部分)分配控制器

/**
 * Display all Assignments for one input name.
 *
 * @param  int  $name
 * @return \Illuminate\Http\Response
 */
public function showAssignmentsByCastName($castName)
{
    if (! Gate::allows('assignment_view')) {
        return abort(403);
    }
    $relations = [
        'anchors' => \App\User::get()->pluck('name', 'id'),
        'casts' =>  \App\Newscast::with('for_station')->get()->pluck('name_input_station', 'id'),
        'assignments' => \App\Assignment::with('cast')->get(),
    ];

dump($relations);

return view('assign.list', compact('castName') +$relations);
}

正如我所预期的，这段代码返回assignments的完整集合。

输出

  Anchor  | Cast Name   | Start Date  |   End Date
 ---------|-------------|-------------|-------------
  Susan   | JNDV-AM     | 01 May 2017 | 31 Dec 2999
  Ed      | JNDV-AM     | 02 May 2017 | 06 May 2017
  Jen     | JNDV-PM     | 01 Apr 2017 | 31 Dec 2999
  Ed      | YXWQ-Sports | 01 Apr 2017 | 28 Apr 2017

我已经尝试过几种方法将作业限制在一个newscastId上，到目前为止都没有成功。

2017年5月1日/assignment/list/JNDV-AM 的预期产量

  Anchor  | Cast Name   | Start Date  |   End Date
 ---------|-------------|-------------|-------------
  Susan   | JNDV-AM     | 01 May 2017 | 31 Dec 2999
  Ed      | JNDV-AM     | 02 May 2017 | 06 May 2017

短期作业是暂时的(isTemp=True).在有效的日子里，它应该列在上面。

2017年5月2日至2017年5月6日的预期产出

  Anchor  | Cast Name   | Start Date  |   End Date
 ---------|-------------|-------------|-------------
  Ed      | JNDV-AM     | 02 May 2017 | 06 May 2017
  Susan   | JNDV-AM     | 01 May 2017 | 31 Dec 2999

我正在修改由管理面板生成器工具生成的代码。在我看来，向所有用户和所有类型查询并不是最有效的方法。我认为，由于我只是在为一个新闻节目寻找当前和未来的任务，所以anchors和casts关系应该被过滤。

基本问题我应该做什么改变

'assignments' => \App\Assignment::with('cast')->get(),

只获得JNDV (newscastId = 1)的赋值？

高级问题，您建议如何更改relations，使其只返回当前和未来的JNDV作业，即今天的作业，并且尽可能少的查询？

Answer 1

下面是我想出的工作代码：

(部分)分配控制器

public function showAssignmentsByCastName($calls, $name_input)
{
    if (!Gate::allows('assignment_view')) {
        return abort(403);
    }

    $castName = strtoupper($calls). "-" . $name_input;
    $station_id = \App\Station::where('calls', $calls)->get();
    $station = \App\Station::findOrFail($station_id);

    $cast_id = \App\Newscast::where([
        ['for_station_id', $station->id],
        ['name_input', $name_input]
    ])->get();
    $cast = \App\Newscast::findOrFail($cast_id);

    $relations = [
        'anchors' => \App\User::get()->pluck('name', 'id'),
        'casts' => \App\Newscast::where('cast_id', $cast->id),
        'assignments' => \App\Assignment::where('cast_id', $cast->id)->get(),
    ];
    return view('assign.list', compact('castName') + $relations);
}

网络路由

Route::get('assign/{calls}-{name_input}', 'AssignmentsController@showAssignmentsByCastName')->name('assign.list');

Answer 2

可以将一个函数传递到with语句中：

'assignments' => \App\Assignment::with(['cast' => function ($query) {
    $query->where('newscastId', '=', 1);
}])->get()