将用户分配到 Laravel 关系中的任何商店？

Question

在 laravel 中，我有 3 张桌子

用户 // 用于身份验证并在登录后创建另一个用户 费用 店铺

我的目的-我希望用户可以注册，并且可以在他们登录时创建另一个用户并且可以根据需要将assign 用户添加到另一个Shop..

只有User in the Same Shop 可以看到他们的Expense..

//我的User表

<pre>
   Schema::create('users', function (Blueprint $table) {
        $table->bigIncrements('id');
        $table->unsignedBigInteger('shop_id')->nullable();
        $table->unsignedBigInteger('user_id')->nullable();
        $table->string('name');
        $table->string('email')->unique();
        $table->timestamp('email_verified_at')->nullable();
        $table->string('password');
        $table->rememberToken();
        $table->timestamps();
    });
</pre>

//我的Expense表

<pre>
    Schema::create('expenses', function (Blueprint $table) {
        $table->bigIncrements('id');
        $table->unsignedBigInteger('user_id');
        $table->date('date');
        $table->string('description');
        $table->double('amount');
        $table->timestamps();
        $table->foreign('user_id')->references('id')->on('users');
    });
</pre>

//我的Shop表

<pre>
     Schema::create('shops', function (Blueprint $table) {
        $table->bigIncrements('id');
        $table->unsignedBigInteger('expense_id')->nullable();
        $table->unsignedBigInteger('user_id');
        $table->string('name');
        $table->string('description');
        $table->timestamps();
        $table->foreign('expense_id')->references('id')->on('expenses');         
        $table->foreign('user_id')->references('id')->on('users');         
    });
</pre>

// 我的用户模型

<pre>
   public function expense()
    {
        return $this->hasMany(\App\Expense::class);
    }


    public function shop()
    {
        return $this->hasMany(\App\Shop::class, 'user_id');
    }
</pre>

// 我的费用模型

<pre>
class Expense extends Model
{
    protected $fillable = ['date', 'description', 'amount', 'user_id', 'shop_id'];

    public function user()
    {
        return $this->belongsTo(\App\User::class);  
    }

}
</pre>

// 我的店铺模型

<pre>
class Shop extends Model
{
    protected $fillable = ['name', 'description', 'expense_id', 'shop_id'];

    public function user()
    {
        return $this->belongsTo(\App\User::class, 'user_id');
    }

}
</pre>

// 费用控制人

<pre>
 public function index(Request $request)
    {
        $expense = Expense::with(['user'])->get();

        return ExpenseResource::collection($expense);

        // dd(auth()->user());
    }

    public function create(Request $request)
    {
        $request->validate([
            'date' => 'required',
            'description' => 'required',
            'amount' => 'required',

        ]);

        $expense = new Expense();
        $expense->user_id = auth()->user()->id;
        $expense->date = $request->date;
        $expense->description = $request->description;
        $expense->amount = $request->amount;
        $expense->save();

        return new ExpenseResource($expense);
    }
</pre>

//在我的用户控制器中

<pre>
public function index()
    {
        $users = User::all();
        $shops = Shop::all();
        return view('user', compact('users', 'shops'));
        // return UserResource::collection($users);
    }

    public function create(Request $request)
    {
        $request->validate([
            'name' => 'required',
            'email' => 'required',
            'password' => 'required',
        ]);

        $user = new user();
        $user->user_id = auth()->user()->id;
        $user->name = $request->name;
        $user->email = $request->email;
        $user->password = bcrypt($request->password);
        $user->save();

        return new UserResource($user);
    }
</pre>

有意义吗？

任何想法，谢谢..

Answer 1

如 cmets 中所述，您需要检查当前的 User 并将返回的 Expense 记录限制为仅 1) 具有 User 和 2) 与 Store 匹配的相同当前User。这可以在单个 whereHas() 子句中完成：

public function index(Request $request) {
  $user = auth()->user(); // If using default `Auth` logic.

  $expenses = Expense::whereHas('user', function($subQuery) use($user){
    return $subQuery->where('shop_id', '=', $user->shop_id);
  })->with(['user'])->get();

  return ExpenseResource::collection($expenses);
}

->whereHas() 所做的是限制获取您的 Expense 模型的查询以尊重您传递的逻辑，在这种情况下仅包括 Expense 具有相同 user 的模型 shop_id作为当前登录的User。

注意：如果当前的User 没有Shop，它可能会返回意外的结果，但您可以保护路由只允许带有Shop 的User 访问它，等等。