Question

我有与多对多关系的物品和用户。我想向用户展示不属于他自己的所有项目，我该如何展示？

编辑：

public function user()
{
  return $this->belongsToMany('App\User', 'interest_user');
}

public function interests()
{
  return $this->belongsToMany('App\Interest', 'interest_user');
}

EDIT2：我更改了代码，但收到了相同的错误消息

public function user()
{
  return $this->belongsToMany(User::class);
}
public function interests()
{
  return $this->belongsToMany(Interest::class);
}

Schema::create('interest_user', function (Blueprint $table) {
        $table->increments('id');
        $table->integer('interest_id')->unsigned();
        $table->foreign('interest_id')->references('id')->on('interests')->onDelete('cascade');
        $table->integer('user_id')->unsigned();
        $table->foreign('user_id')->references('id')->on('users')->onDelete('cascade');
    });


$suggestinterests = Interest::whereHas('user_id', function ($q) use ($user) {
    $q->where('user_id', '!=', $user);
    })->get();

Errormessage：调用未定义的方法App \ Interest :: whereHas（）

EDIT3：这段代码有效！

$suggestinterests =  \App\Interest::whereHas('user', function ($q){
        $q->where('user_id', '!=', auth()->id());
    })->get();

Answer 1

class User
{
    public function items()
    {
       return $this->belongsToMany(Item::class);
    }
}

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


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

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

Schema::create('item_user', function (Blueprint $table) {
    $table->increments('id');

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

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

检索记录：

$availableItems = Item::whereHas('user_id', function ($q) use (Auth::user()->id {
    $q->where('user_id', '!=', Auth::user()->id);
})->get();

这是非常原始的，你可能需要适应它！

Answer 2

我花了一些时间才弄清楚

 $all_items_but_excluded_user_items = DB::table('interests')
         ->select('name',DB::raw('interest_user.interest_id as pivot_interest_id'),
         DB::raw('interest_user.user_id as pivot_user_id'))
         ->join('interest_user','interests.id','=','interest_user.item_id')
         ->where('interest_user.user_id','!=',$user->id)
         ->get();

显示不属于该用户的项目

2 个答案: