Question

我正在尝试获取项目中的上一个和下一个记录（如博客文章）。我无法获取与每个项目的相关的ID。

我可以在我的projectController中的同一查询中这样做吗？

我可以分配并获取每个项目的标题，这些标题是根据标题创建的。

::ProjectController


    public function single($slug){
       
        $project = Project::with('tools', 'categories')->where('slug', $slug)->firstOrFail();
        $prev = Project::where('id', '<', $project->id)->orderBy('id','desc')->first();
        $next = Project::where('id', '>', $project->id)->orderBy('id')->first();
        return view('project-single',compact('project',$project))
                    ->with('prev',$prev)
                    ->with('next',$next);
    }

::SingleProject.blade.php

 @if($prev ==! null)
                <a href="{{ route('project.single', ['project' => $prev]) }}" class="animate">

                  <i class="las la-long-arrow-alt-left"></i>Anterior
                </a> 
                @endif

::Route

Route::get('/proyectos/{slug}', 'ProjectController@single')->name('project.single');

Answer 1

那呢：

$project = Project::with('tools', 'categories')->where('slug', $slug)->firstOrFail();
 

$more = Project::whereIn('id', [$project->id - 1, $project->id + 1])->get();
$prev = $more->where('id', '<', $project->id)->first();
$next = $more->where('id', '>', $project->id)->first();

Answer 2

我解决了这个问题：

::projectController

 public function single($slug){
        $project = Project::with('tools', 'categories')->where('slug', $slug)->firstOrFail();
        $next = Project::where('id', '>', $project->id)->min('id');
        $prev = Project::where('id', '<', $project->id)->max('id');
        
        return view('project-single',compact('project',$project))
                    ->with('prev',Project::find($prev))
                    ->with('next',Project::find($next));
    }

:::view

 @if($prev ==! null)
                <a href="{{ route('project.single', ['slug' => $prev->slug]) }}" class="animate">

                  <i class="las la-long-arrow-alt-left"></i>Anterior
                </a> 
                @endif

如何在Laravel中使用Slug获取下一个和上一个记录

2 个答案: