Question

我用laravel 5.8开发了一个站点，我想使用而不是id。知道我有两个表服务和类型服务的手提袋。服务密钥在服务类型表中是外来的。

我按服务类型检索和显示服务。但在我的网址上会显示类型ID。

我有这个错误

缺少[Route：service] [URI：service / {slug}]的必需参数。（视图：C：\ laragon \ www \ elbi \ resources \ views \ layouts \ partial \ nav.blade.php）（视图：C：\ laragon \ www \ elbi \ resources \ views \ layouts \ partial \ nav.blade php）

表格：

 public function up()
    {
        Schema::create('services', function (Blueprint $table) {
            $table->bigIncrements('id');
            $table->string('name');
            $table->timestamps();
        });
    }


 public function up()
    {
        Schema::create('type_services', function (Blueprint $table) {
            $table->bigIncrements('id');
            $table->unsignedBigInteger('service_id');
            $table->foreign('service_id')->references('id')->on('services')->onDelete('cascade');
            $table->string('name');
            $table->text('description');
            $table->string('image');
            $table->timestamps();
        });
    }

模型

<?php
namespace App;
use Illuminate\Database\Eloquent\Model;

class Service extends Model
{
    public function typeservice()
    {
        return $this->hasMany('App\Typeservice');
    }


}

<?php

namespace App;

use Illuminate\Database\Eloquent\Model;

class TypeService extends Model
{
    /**
     * @return \Illuminate\Database\Eloquent\Relations\BelongsTo
     */
    public function service()
    {
        return $this->belongsTo('App\Service');
    }
    public function getRouteKeyName()
    {
        return 'slug';
    }
}

显示控制器

 public function index()
    {
        $contacts = Contact::all();
        $abouts = About::all();
        $services = Service::with('typeservice')->orderBy('name','asc')->get();
        $typeservices =TypeService::all();
        $makings = Making::all();
        $realisations = Realisation::all();
        $addresses = Address::all();
        return view('index',compact('contacts','services','abouts','typeservices','makings','realisations','addresses'));
    }

 public function service(Service $service)
    {
        $abouts = About::all();
        $services = Service::with(['typeservice'=> function($query){
            $query->where('name');
        }])->orderBy('name','asc')->get();
        $typeservices =$service ->typeservice()->where('service', function ($query) {
            $query->where('name');
        })->paginate(5);;
        $makings = Making::all();
        $realisations = Realisation::all();
        return view('service.service',compact('services','abouts','typeservices','makings','realisations'));
    }

路线

/*controller front */
Route::get('/','FrontController@index')->name('index');
Route::get('/contact','FrontController@send')->name('send');
Route::post('/contact','ContactController@sendMessage');
Route::get('/about','FrontController@about')->name('about');
Route::get('/service/{slug}', 'FrontController@service')->name('service');

视图

   <li class="nav-item submenu dropdown">
            <a href="#" class="nav-link dropdown-toggle" data-toggle="dropdown" role="button" aria-haspopup="true" aria-expanded="false">Services</a>
            <ul class="dropdown-menu">
                @foreach($services as $service)
                <li class="nav-item"><a class="nav-link" href="{{route('service',$service->id)}}">{{$service->name}}</a></li>
                @endforeach
            </ul>
        </li>

我想使用slug而不是id

Answer 1

您可以尝试给Spatie's Laravel-Sluggable包裹放行。

$service = new Service();
$service->name = 'This is a service';
$service->save();

echo $service->slug; // ouputs "this-is-a-service"

Answer 2

您可以通过将slug存储在数据库中并相应地在控制器中修改查询来做到这一点，可以在保存模型之前先设置子弹，如下所示：

$slug = \Illuminate\Support\Str::slug($service->name);
$service->slug = $slug;

您可以在调用save方法或使用Eloquent events-特别是saving事件之前手动放置此逻辑。

使用子弹代替ID

2 个答案: