What is the best practice to handle friendly urls in laravel?

by jose_gulgowski , in category: SEO , a year ago

What is the best practice to handle friendly urls in laravel?

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3 answers

by larry_orn , a year ago

@jose_gulgowski 

In Laravel, the best practice for handling friendly URLs (also known as "pretty URLs" or "clean URLs") is to use the built-in routing system. Laravel's routing system allows you to define routes for your application, which determine how incoming requests are handled and what response is returned.


To create friendly URLs in Laravel, you can define routes using a syntax that includes a URL segment representing the resource you want to access, such as:

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Route::get('/posts/{id}', 'PostController@show');


In this example, the /{id} segment of the URL represents a dynamic parameter that can be used to fetch a specific post from the database. The PostController@show string represents the controller method that will handle the request and return the response.


To generate links to these routes, you can use the route helper function, which generates a URL based on the named route and any parameters it requires:

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


This generates a link to the posts/{id} route, passing in the $post->id value as the id parameter.


By using Laravel's routing system and generating URLs with the route helper function, you can create friendly URLs that are easy to read, remember, and share, while still maintaining a clean and organized codebase.

Member

by julio , 6 months ago

@jose_gulgowski 

Additionally, you can also utilize Laravel's "slug" functionality to create more user-friendly and SEO-friendly URLs. Slugs are a URL-friendly version of a string, typically used in blog post URLs, to represent the title or name of the resource.


To generate a slug in Laravel, you can make use of the str_slug() or slug() helper functions. These functions will convert a given string into a slug format, removing any special characters, converting spaces to hyphens, and making the string lowercase.


For example, you can generate a slug for a blog post title like this:

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$slug = str_slug($post->title);


Then, you can use the generated slug as a parameter in your route definition:

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Route::get('/posts/{slug}', 'PostController@show');


In the PostController@show method, you can fetch the post based on the slug:

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public function show($slug)
{
    $post = Post::where('slug', $slug)->firstOrFail();
    //...
}


This allows you to create more readable and SEO-friendly URLs, such as /posts/my-great-blog-post, while still keeping your route definitions and controller logic clean and organized.

Member

by drew , 6 months ago

@jose_gulgowski 

In addition to the points mentioned above, another best practice for handling friendly URLs in Laravel is to use route resource binding. This feature allows you to bind a model instance to a route parameter automatically.


To use route resource binding, you can define a model binding in the RouteServiceProvider's boot() method. For example, if you want to bind the "Post" model to the "slug" parameter in your routes, you can add the following code:


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public function boot() { Route::bind('slug', function ($value) { return Post::where('slug', $value)->firstOrFail(); }); }


With this binding, Laravel will automatically fetch the Post model instance based on the "slug" parameter value in your route definitions. For example, you can define your route like this:


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Route::get('/posts/{slug}', 'PostController@show');


In the PostController@show method, you can simply type-hint the Post model in the method's signature:


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public function show(Post $slug) { // $slug represents the Post model instance // ... }


By using route resource binding, you can easily retrieve the model instance based on the friendly URL parameter, eliminating the need for manual retrieval and reducing code duplication.


Remember to also add the necessary exception handling to gracefully handle scenarios when the binding fails to find a matching model instance. You can use the firstOrFail() method or the findOrFail() method to throw an exception if the model cannot be found.