獲取ID爲1的設置？ Laravel5.4 *

Question

我有以下表格：

用途表：

Schema::create('users', function (Blueprint $table) { 
     $table->increments('id'); 
     $table->string('firstname'); 
     $table->string('surename'); 
     $table->string('address')->nullable(); 
     $table->string('city')->nullable(); 
     $table->string('country')->nullable(); 
     $table->string('postcode')->nullable(); 
     $table->string('email')->unique(); 
     $table->string('password'); 
     $table->rememberToken(); 
     $table->string('facebook_id')->nullable(); 
     $table->string('linkedin_id')->nullable(); 
     $table->string('twitter_id')->nullable(); 
     $table->timestamps(); 
    }); 

設置表：

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

Setting_user（透視表）：

Schema::create('setting_user', function (Blueprint $table) { 
     $table->integer('user_id')->unsigned(); 
     $table->foreign('user_id')->references('id')->on('users'); 
     $table->integer('setting_id')->unsigned(); 
     $table->foreign('setting_id')->references('id')->on('settings'); 
     $table->timestamps(); 
    }); 

分類表：

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

Setting_category（透視表）：

Schema::create('setting_category', function (Blueprint $table) { 
     $table->integer('category_id')->unsigned(); 
     $table->foreign('category_id')->references('id')->on('categories'); 
     $table->integer('setting_id')->unsigned(); 
     $table->foreign('setting_id')->references('id')->on('settings'); 
     $table->timestamps(); 
    }); 

在我的用戶模型，這個我有一個belongsToMany關係：

public function settings() 
{ 
    return $this->belongsToMany(Setting::class, 'setting_user', 'user_id','setting_id')->with('category'); 
} 

在我的設置模式：

保護$表= 「設置」;

public function category() 
{ 
    return $this->hasMany(Categories::class,'id'); 
} 

當我訪問它，像這樣：

public function settings() 
{ 
    $user = User::find(1); 
    return $user->settings()->with('category')->first(); 
} 

這是結果我得到：

{"id":1,"value":"87658675.jpg","created_at":"2017-07-04 00:00:00","updated_at":null,"pivot":{"user_id":1,"setting_id":1},"category":[{"id":1,"name":"avatar","created_at":"2017-07-06 00:00:00","updated_at":null}]} 

如何我類別 - > id來獲取只有設置== 1 ？

非常感謝

Answer 1

如果你想只對指定的用戶和指定的類別設置，使用whereHas()：

public function settings() 
{ 
    $categoryId = 1; 
    $userId = 1; 

    return Settings::whereHas('categories', function ($q) use ($categoryId) { 
      $q->where('id', $categoryId); 
     }) 
     ->whereHas('users', function ($q) use ($userId) { 
      $q->where('id', $userId); 
     }) 
     ->get(); 
} 

爲了使這項工作，解決您的關係。貌似所有的關係應該是belongsToMany()，而不是hasMany()

Answer 2

我已經在我的設置模式改變了關係到：

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

public function categories() 
{ 
    return $this->belongsToMany(Categories::class,'setting_category','category_id','setting_id'); 
} 

我仍然在我的用戶模型belongsToMany關係：

public function settings() 
{ 
    return $this->belongsToMany(Setting::class, 'setting_user', 'user_id','setting_id'); 
} 

而且我創建了另一種方法來提取id爲「1」的頭像，如下所示：

/** 
* Get user's avatar. 
*/ 
public function getAvatarAttribute() { 

    $categoryId = 1; 
    $userId = Auth::user()->id; 

    $avatar = $this->settings()->whereHas('categories', function ($q) use ($categoryId) { 
     $q->where('id', $categoryId); 
    })->whereHas('users', function ($q) use ($userId) { 
     $q->where('id', $userId); 
    })->first(); 

    if(!$avatar) 
    { 
     return "default-avatar.jpg"; 
    } 

    return $avatar->value; 

}