Laravel 4关系排序一对多计数急切加载

Question

不确定这是否可能。我可以对当前表格艺术家（标题、正文等）中的所有内容进行排序，但还没有想出将其排序到与之相关的表格中。我想根据他/她拥有 DESC、ASC 的歌曲总数对艺术家进行排序。

如何动态排序艺术家的歌曲总数？

admin/artists?sortby=songs&order=asc

Screenshot

我的桌子

艺术家：id、名字、身体、蛞蝓

歌曲：id、title、body、slug、hits、artist_id

在此之前，我在控制器中使用了预加载，Artist::with('songs)... 然后在我可以在 foreach 循环中使用的视图中，count($artist->songs) 用于总歌曲。然后我找到了一种更清洁的方法 http://laravel.io/forum/05-03-2014-eloquent-get-count-relation

艺人模特

public function songs()
    {
        return $this->hasMany('Song');
    }

public function songsCountRelation()
{
    return $this->hasOne('Song')->selectRaw('artist_id, count("id") as count')->groupBy('artist_id');
}

public function getSongsCountAttribute()
{
    return $this->songsCountRelation->count;
}

歌曲模型

public function artist()
{
    return $this->belongsTo('Artist');
}   

控制器

public function index()
{
    // Sortby and order variables from url 
    $this->data['sortby'] = Input::get('sortby');
    $this->data['order'] = Input::get('order');

    // If the sortby and order exists in url, fetch the data and order it accordingly
    if ($this->data['sortby'] && $this->data['order'])
    {
        // Order the data based on keys from url
        $this->data['artists'] = Artist::with('songsCountRelation')->orderBy($this->data['sortby'], $this->data['order'])->get(['id', 'name', 'body', 'slug']);

    }
    else
    {
        $this->data['artists'] = Artist::with('songsCountRelation')->get(['id', 'name', 'body', 'slug']);
    }

    $this->layout->content = View::make('admin.artists.index', $this->data);
}

索引视图

<th>
    {{--Sort by slug--}}
    @if ($sortby == 'slug' && $order == 'asc')

        {{ link_to_route('admin.artists.index', 'Slug', ['sortby' => 'slug', 'order' => 'desc']) }}
    @else
        {{ link_to_route('admin.artists.index', 'Slug', ['sortby' => 'slug', 'order' => 'asc']) }}
    @endif
</th>
<th>
    {{--Sort by total songs--}}
    @if ($sortby == 'songs' && $order == 'asc')
        {{ link_to_route('admin.artists.index', 'Songs', ['sortby' => 'songs', 'order' => 'desc']) }}
    @else
        {{ link_to_route('admin.artists.index', 'Songs', ['sortby' => 'songs', 'order' => 'asc']) }}
    @endif
</th>
...¨
@foreach($artists as $artist)
    <tr>
...
        <td>{{ $artist->songsCount; }}</td>
...     
    </tr>

@endforeach

提前致谢！

Answer 1

你必须加入表格。

Artist::join('songs', ...)
  ->orderByRaw('count(songs.*) asc')
  ->select('artists.*')
  ->get();