Laravel 4 Eager Loading n+1 问题

Question

我整天都在解决急切加载/n+1 问题，研究、阅读和观看了有关此问题的教程，但尚未解决。我已经为模型设置了关系，但是当我使用辅助函数传入数据时，我遇到了这个 n+1 问题。 我想从 url site.com/Artist/songs 中获取一个艺术家的名字并获取它的所有歌曲并显示一个这样的 url。

site.com/$Artist/songs/$id

我的艺术家/index.blade.php 视图看起来像这样http://i61.tinypic.com/2nqzatk.jpg 我不确定我在这里缺少什么。

提前致谢！

我的桌子

歌曲 id、标题、正文、slug、点击数、artist_id、created_at、updated_at

艺术家 id, name, body, created_at, updated_at

routes.php

Event::listen('illuminate.query', function($query)
{
    var_dump($query);
});
...
Route::get('{artist}/songs', 'ArtistsController@index');
Route::get('{artist}/songs/{id}', ['as' => 'artist.songs.show', 'uses' => 'ArtistsController@show']);

型号：Song.php

class Song extends Eloquent {
    protected $guarded = ['id'];

    /**
     * Setting up relationship for the artist model for easier usage
     *
     */
    public function artist()
    {
        return $this->belongsTo('Artist');
    }

    // Override find method
    public static function find($id, $name = null)
    {
        $song = static::with('artist')->find($id);

        // If the song is found
        if ($song)
        {
            // If the song doesn't belong to that artist, throw an exception which will redirect to home, defined in global.php
            if ($name and $song->artist->name !== $name)
            {
                throw new Illuminate\Database\Eloquent\ModelNotFoundException;
            }

            return $song;
        }
        // If the song is not found, throw an exception which will redirect to home, defined in global.php
        else
        {
            throw new Illuminate\Database\Eloquent\ModelNotFoundException;
        }

    }

    // Get songs from artist
    public static function byArtist($name)
    {
        return Artist::byName($name)->songs;
    }

}

模型艺术家.php

class Artist extends Eloquent {
    protected $fillable = [];

    /**
     * Setting up relationship with the song model for easier usage
     * $artist->songs;
     */
    public function songs()
    {
        return $this->hasMany('Song');
    }

    // Get artist by name
    public static function byName($name)
    {
        return static::whereName($name)->first();
    }

}

控制器：ArtistsController.php

class ArtistsController extends BaseController {

    // Set default layout for this controller
    protected $layout = 'layouts.master';

    /**
     * Display a listing of the resource.
     * GET /artists
     *
     * @return Response
     */
    public function index($name)
    {
        $this->data['songs'] = Song::byArtist($name);

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

helpers.php

function link_to_artist_song(Song $song)
{
    return link_to_route('artist.songs.show', $song->title, [$song->artist->name, $song->id]);
}

艺术家的索引视图 艺术家/index.blade.php http://i61.tinypic.com/2nqzatk.jpg

@extends('layouts.master')

@section('content')

    @if(isset($songs))
        <h1>All Songs</h1>

        <ul class="list-group">
        @foreach($songs as $song)
            <li class="list-group-item">{{ link_to_artist_song($song) }}</li>
        @endforeach
        </ul>
    @endif
@stop

Answer 1

当您有很多歌曲和很多艺术家时，就会出现 n+1 问题。它必须获取所有歌曲（1 个查询），然后为每首歌曲获取艺术家（n 个查询）。

在这种情况下，您已经知道这首歌，所以这是一个查询，然后您需要该歌曲的所有艺术家，这只是一个额外的查询。

n+1 问题只有在您尝试查找某一类型的歌曲时才会发挥作用，例如，这可能会返回许多歌曲。然后，对于每首歌曲，您必须进行额外的查询以获取该歌曲的艺术家。这将是急切加载最有用的地方。

$song = Song::find($id);
$artist = $song->artist;

Answer 2

您从不急于加载任何内容，这就是您可能面临 n+1 问题的原因。 如果我猜对了，你在这里的代码有什么难的地方，你想要给定艺术家的所有歌曲，其中 $name 来自 url，对吧？

所以这里有你需要的一切：

// controller
public function index($name)
{
    // with('songs') is eager loading related songs for you
    $this->data['artist'] = Artist::with('songs')->whereName($name)->first();

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

// the problem of your queries is in the helper:
function link_to_artist_song(Song $song)
{
    return link_to_route('artist.songs.show', $song->title, [
       $song->artist->name, // this is calling db query for each song to retrieve its artist (despite it is always the same)
       $song->id]);
}

// so instead use this in your view
@foreach($artist->songs as $song)
   <li class="list-group-item">
     {{ link_to_route('artist.songs.show', $song->title, [$artist->name, $song->id]) }}
   </li>
@endforeach

Answer 3

每次你抓取一首歌，然后做$song->artist，它都会做一个查询。

您也可以使用Query scopes。

class Song extends Eloquent {

    public function scopeByArtist($query, $name) {
        return Artist::whereName($name)->first()->songs(); //->with('artist');
    }
}

像这样，艺术家已经加载完毕。

查询：

$songs = Song::byArtist($name)->get();