Question

我在web.php中有这些路线：

Route::get('/threads/create','ThreadForumController@create');
Route::get('/threads/{channel}','ThreadForumController@index');
Route::post('/threads','ThreadForumController@store');
Route::get('/threads/{channel}/{thread}','ThreadForumController@show')->name('threads.show');
Route::post('/threads/{channel}/replies','ReplyController@store');
Route::get('/threads','ThreadForumController@index');

当我进入浏览器/ thread时，我必须得到所有线程，当我去/ threads / php时我必须获得specefic线程，但我的控制器中总是有$ channel值：

public function index(Channel $channel)
    {
        $threads = $channel->exists() ? $channel->threads()->latest()->get() : ThreadForum::all();

        return view('threads.index', compact('threads'));
    }

我该如何解决？

Answer 1

您可以将两条路线合并为一条：

Route::get('/threads/{channel}','ThreadForumController@index');
Route::get('/threads','ThreadForumController@index');

可以改为：

Route::get('/threads/{channel?}','ThreadForumController@index');

然后index方法可写如下：

public function index($channel == null) {
    if ($channel) {
        //I always go in this way
        $threads = Channel::findOrFail($channel)->threads()->latest()->get();
    } else {
        $threads = ThreadForum::all();
    }
    return view('threads.index', compact('threads'));
}

因此，您可以使用一个路线和一个方法同时服务。

Answer 2

有几种方法，一种是这样的：

 public function index(Channel $channel)
{
    $threads = Thread::latest();

    if(request('channel')) {
        $threads = $channel->threads()->latest();
    }

    $threads = $threads->get();

    return view('threads.index', compact('threads'));
}

如果您想要在路线中使用slugs（'php'）而不是id，则必须设置getRouteKeyName。例如，在频道模型中：

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

public function threads()
{
    return $this->hasMany(Thread::class);
}

Laravel两种方法的路线

2 个答案: