在Laravel 5.2中使用雄辯而不是查詢

Question

我提出了一個查詢，但我並不滿意，因爲我想用Eloquent來做！

下面是該查詢：

Tournament::leftJoin('category_tournament', 'category_tournament.tournament_id', '=', 'tournament.id') 
     ->leftJoin('category_tournament_user', 'category_tournament_user.category_tournament_id', '=', 'category_tournament.id') 
     ->where('category_tournament_user.user_id', '=',$this->id) 
     ->select('tournament.*') 
     ->distinct() 
     ->get(); 

遷移：

Schema::create('tournaments', function(Blueprint $table) { 
    $table->increments('id'); 
    $table->string('name'); 
    $table->timestamps(); 
}); 

Schema::create('categories', function(Blueprint $table) { 
    $table->increments('id'); 
    $table->string('name'); 
    $table->timestamps(); 
}); 

Schema::create('category_tournament', function(Blueprint $table) { 
    $table->increments('id'); 
    $table->integer('category_id'); 
    $table->integer('tournament_id'); 
    $table->timestamps(); 

    $table->foreign('tournament_id') 
      ->references('id') 
      ->on('tournament') 
      ->onDelete('cascade'); 

     $table->foreign('category_id') 
      ->references('id') 
      ->on('category') 
      ->onDelete('cascade'); 
}); 

Schema::create('category_tournament_user', function (Blueprint $table) { 
     $table->increments('id'); 
     $table->integer('category_tournament_id')->unsigned()->index(); 
     $table->foreign('category_tournament_id') 
      ->references('id') 
      ->on('category_tournament') 
      ->onDelete('cascade'); 


     $table->integer('user_id')->unsigned()->index(); 
     $table->foreign('user_id') 
      ->references('id') 
      ->on('users') 
      ->onDelete('cascade'); 

     $table->unique(array('category_tournament_id', 'user_id')); 

     $table->boolean('confirmed'); 

     $table->timestamps(); 
     $table->softDeletes(); 
     $table->engine = 'InnoDB'; 


    }); 

模型

class Tournament extends Model 
{ 
    public function categories() 
    { 
     return $this->belongsToMany(Category::class); 
    } 

    public function categoryTournaments() 
    { 
     return $this->hasMany(CategoryTournament::class); 
    } 
} 

class Category extends Model 
{ 
    public function tournaments() 
    { 
     return $this->belongsToMany(Tournament::class); 
    } 

    public function categoryTournament() 
    { 
     return $this->hasMany(CategoryTournament::class); 
    } 
} 

class CategoryTournament extends Model 
{ 
    protected $table = 'category_tournament'; 

    public function category() 
    { 
     return $this->belongsTo(Category::class); 
    } 

    public function tournament() 
    { 
     return $this->belongsTo(Tournament::class); 
    } 

    public function users() 
    { 
     return $this->belongsToMany(User::class, 'category_tournament_user'); 
    } 
} 

class User extends Authenticatable 
{ 
    public function categoryTournaments() 
    { 
     return $this->belongsToMany(CategoryTournament::class, 'category_tournament_user'); 
    } 
} 

的查詢工作，我只是想知道我應該怎麼用雄辯這樣做是因爲我無法自己做：（

任何想法如何做？

Answer 1

首先，你必須外鍵添加到您的遷移表+，使他們無符號的，因爲你可能永遠不會有任何負面的ID是非常有用：

$table->integer('tournament_id')->unsigned(); 
$table->foreign('tournament_id')->references('id')->on('tournaments'); 

既然你已經爲你的模型中的關係你應該能夠使用雄辯來獲得內容。

你能堅持到這個頁面，進一步瞭解如何雄辯： https://laravel.com/docs/5.0/eloquent

應該是這樣的：

$tournaments = Tournament::with('category_tournament') 
    ->with('category_tournament_user')  
    ->where('category_tournament_user.user_id', '=',$this->id) 
    ->distinct() 
    ->get(); 

您可以使用$比賽內容算賬：

$tournaments->category_tournament->name; 

希望我沒有錯過任何關鍵的步驟，但我認爲它應該通過做這些改變來工作。

編輯：

https://laravel.com/docs/5.1/eloquent-relationships#eager-loading

當訪問作爲屬性口才關係，關係數據是 「延遲加載」。這意味着關係數據在第一次訪問該屬性之前並未實際加載。但是，Eloquent可以在查詢父模型時「加載」關係。急切的加載減輕了N + 1查詢問題。要說明N + 1查詢問題，請考慮與作者有關​​的Book模型作者：