如何在Laravel 5.6中显示多对多关系中的最后一条记录

Question

I have multiple images in my uploads table with related to vehicle_id, as foreign key like this, 我的上传表中有多张图片，其中包含与vehicle_id相关的内容，就像这样的外键，

uploads table
id          fileName        vehicle_id
1            1.jpg               1
2            2.jpg               1
3            3.jpg               1
4            4.jpg               1
5            28.png              2
6            28.png              2
7            29.png              2
8            30.png              3
9            31.png              3
10           56.png              3

The vehicle table have many to many relationship with the image table and data grap using eager loader in VehicleController, 车辆表格与图像表格和使用VehicleController中的eager loader进行的数据抓取有很多关系，

$vehicles = Vehicle::with('uploads')->get();
        return view('vechicles.index')->withVehicles($vehicles);

Now these images are showing in the vehicles/index.blade.php file 现在这些图像显示在Vehicles / index.blade.php文件中

 @foreach($vehicle->uploads as $upload)

         <tr>
                        <td><a href="{{route('vechicles.show',$vehicle->id)}}"><img src="/images/{{ $upload->resized_name }}"></a></td>

                    </tr>

        @endforeach

My problem is occurred now, in this way, I can show all images in the table whith related to proper vehicle_id but, I need only show one image (to matching vehicle_id ) like thumbnail to above line. 我的问题现在发生了，用这种方式，我可以在表中显示与正确的vehicle_id相关的所有图像，但是，我只需要在上面一行显示一个图像（与匹配的vehicle_id相匹配），如缩略图。 Then how can I configure this? 那我该如何配置呢？ actually I need only one image to print in index.blade.php to related in each vehicle_id, as an example here are 4 images with related to vehicle_id 1. but I need only print decending oder one image to print and same to vehicle_id 2 and etc.... how can do this? 实际上，我只需要在index.blade.php中打印一张要在每个vehicle_id中关联的图像，例如，这里有4张与vehicle_id 1相关的图像，但是我只需要打印下降的一张图像就可以打印，并且与vehicle_id 2相同，并且等等...该怎么做？

Updated User Model 更新的用户模型

class Upload extends Model
{
    public function vehicle()
    {
        return $this->belongsTo(Vehicle::class);
    }
}

updated Vehicle Model 更新的车辆型号

 class Vehicle {
    public function thumbnail() {
        return $this->hasOne(Upload::class)->orderByDesc('id');
    }
}

$vehicles = Vehicle::with('thumbnail')->get();

current Blade file 当前的刀片文件

 @if($vehicles)
@foreach($vehicles as $vehicle)
{{$vehicle->district}}
{{$vehicle->town}}
{{$vehicle->brand}}
{{$vehicle->model}}

<hr>

 <tr>
  @foreach($vehicle->thumbnail as $aaa)
                    <img src="/images/{{ $aaa->resized_name }}">
                    @endforeach

                </tr>


@endforeach
@endif

Answer 1

@Jonas Staudenmeier solution is the cleanest. @Jonas Staudenmeier解决方案是最干净的。 However, if you cannot for any reason update the model, you can use this in any of your blade template. 但是，如果由于某种原因无法更新模型，则可以在任何刀片服务器模板中使用它。

app/Http/Controllers/VehiculeController.php app / Http / Controllers / VehiculeController.php

<?php
    namespace App\Http\Controllers;

    use Illuminate\Http\Request;
    use App\Vehicule;

    class VehiculeController extends Controller {
        public function index() {
            $vehicules = Vehicule::with('uploads')->get();

            return view('vehicule.index')->withVehicules($vehicules);
        }
    }
?>

resources/views/vehicule/index.blade.php 资源/视图/车辆/index.blade.php

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <meta http-equiv="X-UA-Compatible" content="ie=edge">
    <title>Document</title>
</head>
<body>
    <table>
        <tbody>
            @forelse( $vehicules as $vehicule )
                <td>
                    @if( $vehicule->uploads->count() > 0 )
                        <a href="{{ route('vehicules.show', $vehicule->id) }}">
                            @php
                                $upload = $vehicule->uploads->sortByDesc('id')->first();
                            @endphp
                            <img src="{{ $upload->resized_name }}" alt="{{ $upload->fileName }}" />
                        </a>    
                    @else
                        This vehicule does not have any images attached.
                    @endif
                </td>
            @empty
                <td>No vehicules to display.</td>
            @endforelse
        </tbody>
    </table>
</body>
</html>

Here we use Laravel Collection ->first() method to get the first method, but do not forget (this is done in the example) to check if there is any uploads (or any relationship in a general manner) when you get the first element. 在这里，我们使用Laravel Collection- ->first()方法获取第一个方法，但是不要忘记（在示例中已完成）在获取第一个方法时检查是否存在任何上传（或以一般方式存在任何关系）元件。

When you fetch relations of a model, all Laravel Collection methods are available until you use a finisher ( first , all , ...). 当您获取模型的关系时，所有的Laravel Collection方法都可用，直到您使用了修整器（ first ， all ，...）。

Answer 2

Please try latest method in relationship to get the latest model. 请尝试关系中的最新方法以获取最新模型。

class Vehicle {
    public function thumbnail() {
        return $this->hasOne(Upload::class)->latest();
    }
}

如何在Laravel 5.6中显示多对多关系中的最后一条记录

问题描述

2 个解决方案

解决方案1
0 已采纳 2018-07-13 16:08:08

解决方案2
0 2018-07-13 16:23:05

如何在Laravel 5.6中显示多对多关系中的最后一条记录

问题描述

2 个解决方案

解决方案1 0 已采纳 2018-07-13 16:08:08

解决方案2 0 2018-07-13 16:23:05

解决方案1
0 已采纳 2018-07-13 16:08:08

解决方案2
0 2018-07-13 16:23:05