Question

我试图创建一个简单的Laravel 5应用程序用于测试目的，我仍然在努力通过模型关系显示所需的数据。当我通过工匠修补匠测试模型关系时，一切正常。在我的索引视图中，一切都很好......直到我使用第二个链接模型，这给了我主题中的错误。

以下是我的表格（迁移）：

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

Schema::create('products', function(Blueprint $table)
{
    $table->increments('id');
    $table->integer('order_id')->unsigned();
    $table->timestamps();
    $table->string('name');
    $table->string('sn');
});

Schema::create('services', function(Blueprint $table)
{
    $table->increments('id');
    $table->integer('order_id')->unsigned();
    $table->timestamps();
    $table->string('name');
    $table->string('type');
});

以下是我的3个型号：

//文件：App \ Order.php

<?php namespace App;

use Illuminate\Database\Eloquent\Model;

class Order extends Model {

    public function products()
    {
        return $this->hasMany('App\Product');
    }

    public function services()
    {
        return $this->hasMany('App\Service');
    }
}

//文件：App \ Product.php

<?php namespace App;

use Illuminate\Database\Eloquent\Model;

class Product extends Model {

    public function order()
    {
        return $this->belongsTo('App\Order');
    }
}

//文件：App \ Service.php

<?php namespace App;

use Illuminate\Database\Eloquent\Model;

class Service extends Model {

    public function order()
    {
        return $this->belongsTo('App\Order');
    }
}

//我的简单控制器OrdersController.php

public function index()
{
    $orders = Order::all();
    return view('orders.index', compact('orders'));
}

最后我的观点＆＃34; /resources/views/orders/index.blade.php"

<ul>
    @foreach( $orders as $order )
        <li>
            Order# {{ $order->id }} -             //this line is OK
            {{$order->products->first()->name}} - //this line is OK 
            {{$order->services->first()->name}} - //as soon as I add this line, I get the error! 
        </li>
    @endforeach
</ul>

=＆GT;难道我做错了什么？是不是可以从2＆＃34;外部＆＃34;中检索数据。同一视图中的模型？非常感谢你的帮助！

Answer 1

每个订单是否至少分配了一项服务？我的猜测不是，在这种情况下你应该把它包装在if语句中：

@foreach( $dossiers as $dossier )
    <li>
    Order# {{ $order->id }}
    @if($product = $order->products->first())
        {{ $product->name }}
    @endif
    @if($service = $order->services->first())
        {{ $service->name }}
    @endif
    </li>
@endforeach

Laravel 5：从2个链接模型中获取数据时“尝试获取非对象属性”

1 个答案: