Laravel 5.5仅加载当前登录用户的数据

问题描述 投票:2回答:1

嗨,我是laravel的新手,但我想为当前登录的用户加载所有预订。

我试过这样做

   //check if user is logged in
     if ($user = Auth::user()) {  
       //get only the bookings for the currently logged in user
        $allProducts =Booking::where('client', Auth::user()->name)->where('name', $name)->first();
        //store the bookings in a products variable
        $products = json_decode(json_encode($allProducts));
      //Loop through the products:
        foreach ($products as $key => $val) {
            //get the name of the service by matching it's id in the service model to the service column in the products
            $service_name = Service::where(['id' => $val->service])->first();
      //get the charge amount of the service by matching it's id in the Charge model to the charge column in the products
            $service_fee = Charge::where(['id' => $val->charge])->first();
          //get the status of the service by matching it's id in the status model to the status column in the products
            $service_status = Status::where(['id' => $val->status])->first();
            $products[$key]->service_name = $service_name->name;
            $products[$key]->service_fee = $service_fee->total;
            $products[$key]->service_status = $service_status->name;
        }
        return view('client.booking.view_bookings')->with(compact('products'));
    }
    return view('/login');
}

但这给了我一个错误:未定义的变量:行上的名称

       $allProducts =Booking::where('client', Auth::user()->name)->where('name', $name)->first();

我能做错什么?以及如何解决它只显示所需的数据

laravel-5.5
1个回答
1
投票

我试图了解你在做什么但没有成功,但是从你在评论中的解释,我想我知道你想做什么。

既然你说这个代码适合你,除了它给你数据库中所有数据的结果,无论登录用户如何

    $allProducts = Booking::get();

这是因为它创建了一个选择数据库中所有数据的查询。

您需要的是在语句中添加where子句。要做到这一点,只需将其添加到上面的代码行

    where('client', Auth::user()->name)

它将仅返回包含客户端列的数据等于当前登录用户的名称。

因此整个代码行变成了;

    $allProducts = Booking::get()->where('client', Auth::user()->name);

或者,您可以使用过滤器

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