检查模型是否存在，如果在 Laravel 中没有找到则继续路由

Question

I have two models that I don't want to have a prefix in front of its URLs. Eg Users and Posts

If I have a URL https://example.com/title-of-the-post and https://example.com/username

I will do something like this in the web.php routes file:

// User
Route::get('{slug}', function ($slug) {
    $user = User::whereSlug($slug)->get();
    return view('users.show', $user);
});

// Post
Route::get('{slug}', function ($slug) {
    $post = Post::whereSlug($slug)->get();
    return view('posts.show', $user);
});

Now the issue I am facing is that the first route is entered and will never reach the second even if there is no model with a matching slug.

How can I exit to the next route (Post) if $user is not found?

Note: I have tried many different exit strategies but none seem to work.

return;
return false;
return null;
// and return nothing

Thanks in advance!

UPDATE :

Another issue is that if I have other resource routes they too get blocked by the first route.

Eg If I have Route::resource('cars', 'CarController') it generates a /cars path which matches the {slug} and is also being blocked by first User route.

Answer 1

I think you already got the idea but I ,kind of, have a similar setup in my application and in my particular case, I also needed to be able to catch multi-segments routes.

So here's how I did it. For example, the last route in my web.php is the following.

Route::get('{catchall}', 'SlugRoutesController@route')->where('catchall', '.*');

The where class ->where('catchall', '.*'); ensures that we are also able to catch slugs that have multiple segments.

For example, the following routes will all be matched:

/blog/this-is-an-article
/user/mozammil/articles

Then, in my SlugRoutesController , I am able to inject my other Controller dependencies.

<?php

namespace App\Http\Controllers;

use App\Post; 
use App\User; 
use PostController;
use UserController; 
use Illuminate\Http\Request;

class HomeController extends Controller
{
    private $postController;
    private $userController;

    /**
     * Create a new controller instance.
     *
     * @return void
     */
    public function __construct(UserController $userController, PostController $postController)
    {
        $this->userController = $userController; 
        $this->postController = $postController; 
    }

    public function route(Request $request, string $slug)
    {
        $post = Post::where('slug', $slug)->first(); 
        if(post) {
            return $this->postController->index($request); 
        }

        $user = User::where('slug', $slug)->first(); 
        if($user) {
            return $this->userController->index($request); 
        }

        abort(404);
    }
}

My actual controller is a bit more complex than that, but you get the idea.

Answer 2

You should check both of them in one route :

Route::get('{slug}', function ($slug)
{
    $user = User::whereSlug($slug)->first();

    if ($user)
    {
        return view('users.show', $user);
    }
    else
    {
        $post = Post::whereSlug($slug)->first();

        if ($post)
        {
            return view('posts.show', $post);
        }
        else
        {
            abort(404);
        }
    }
});

However, it can be more cleaner. But the concept is there.

Answer 3

Not sure if this is best practice but here is what was done to accomplish what I needed.

Created a function on the Post model that checks if the slug is calling a post. It uses both a regex and database lookup.

public static function isRequestedPathAPost() {
    return !preg_match('/[^\w\d\-\_]+/', \Request::path()) &&
        Post::whereSlug(\Request::path())->exists();
}

Then I wrap the Route in a if statement like this.

if (\App\Models\Post::isRequestedPathAPost()) {
    Route::get('{slug}', 'PostController@show');
}

Now the route is only used if it actually exist. You can put this at the bottom of the route file to reduce unnecessary lookups to the database.