laravel 5.8：slug 404 not found

Question

我想使用slug，但是当我单击并跳转到特定帖子时，404找不到。

URL运作良好，所以我不明白为什么我看不到结果。

web.php

Route::get('results/{post}', 'ResultsController@show')->name('posts.show');

post.php

public function getRouteKeyName()
{
    return 'slug';
}

ResultsController.php

public function show(Post $post)
{
    $recommended_posts = Post::latest()
                        ->whereDate('date','>',date('Y-m-d'))
                        ->where('category_id','=',$post->category_id)
                        ->where('id','!=',$post->id)
                        ->limit(7)
                        ->get();


    $posts['particular_post'] = $post;
    $posts['recommended_posts'] = $recommended_posts;

    return view('posts.show',compact('posts'));
}

表格

Schema::create('posts', function (Blueprint $table) {
        $table->bigIncrements('id');
        $table->string('image');
        $table->unsignedBigInteger('category_id');
        $table->string('title');
        $table->string('slug');
        $table->string('place');
        $table->string('map');
        $table->date('date');
        $table->string('organizer');
        $table->string('organizer_link');
        $table->timestamp('published_at')->nullable();
        $table->text('description');
        $table->timestamps();
    });

PostsController.php

 public function store(CreatePostsRequest $request)
{
    //upload the image to strage
    //dd($request->image->store('posts'));
    $image = $request->image->store('posts');

    //create the posts
    $post = Post::create([
        'image' => $image,
        'category_id' => $request->category,
        'title' => $request->title,
        'slug' => str_slug($request->title),
        'place' => $request->place,
        'map' => $request->map,
        'date' => $request->date,
        'organizer' => $request->organizer,
        'organizer_link' => $request->organizer_link,
        'published_at' => $request->published_at,
        'description' => $request->description
    ]);

result.blade.php

<a href="{{ route('posts.show', [$post->id,$post->slug]) }}" class="title-link">{{ str_limit($post->title, 20) }}</a>

Answer 1

这称为Route Model Binding。您的路线应类似于：

Route::get('results/{post}', 'ResultsController@show')->name('posts.show');

然后根据您的控制器

public function show(Post $post)
{
    $slug = $post->slug;
    $recommended_posts = Post::latest()
                        ->whereDate('date','>',date('Y-m-d'))
                        ->where('category_id','=',$post->category_id)
                        ->where('id','!=',$post->id)
                        ->limit(7)
                        ->get();


    $posts['particular_post'] = $post;
    $posts['recommended_posts'] = $recommended_posts;

    return view('posts.show',compact('posts'));
}

现在，您需要做的更改在这里。

<a href="{{ route('posts.show', ['id'=>$post->id]) }}" class="title-link">{{ str_limit($post->title, 20) }}</a>

Answer 2

您已定义模型以将slug键用于隐式路由模型绑定。您定义的路由results/{post}带有1个参数post。您正在向路由帮助程序传递一个ID和一个标签，这使它使用ID作为参数：

route('posts.show', [$post->id, $post->slug])

您无需为此路线传递帖子的ID，您希望对参数使用slug：

route('posts.show', $post->slug);
// or
route('posts.show', ['post' => $post->slug]);