Right way to build a link in laravel 5.3

问题:

Im trying to build a dynamic link with a view page (blade) with Laravel 5.3.

My approach is:

<a href=" {{ URL::to('articles') }}/{{ $article->id}}/edit">Edit></a>  

that will output the right url with my base url and some other slug: http://mydomain/articles/23/edit
Where "23" is my article's id.

This works but I wonder if there is a cleaner way to do that?

many thanks

回答1:

You can use named routes for this

// Your route file
URL::get('articles/{articleId}/edit', 'ArticlesController@edit')->name('articles.edit');

//Your view
<a href="{{ URL::route('articles.edit', $article->id) }}">Edit</a>

Much more cleaner IMO

回答2:

You can use named routes for cleaner in code

In your app/Http/routes.php (In case of laravel 5, laravel 5.1, laravel 5.2) or app/routes/web.php (In case of laravel 5.3)

Define route

Route::get('articles/{id}/edit',[
             'as'   =>'articles.edit',
             'uses' =>'YourController@yourMethod'
            ]);

In Your view page (blade) use

<a href="{{ route('articles.edit',$article->id) }}">Edit</a>

One benefits of using named routes is if you change the url of route in future then you don't need to change the href in view (in your case)

回答3:

You can try with this

<a href="{{ url('/articles/edit',$article->id) }}"><i class="fa fa-fw fa-edit"></i></a>

and your route.php file

Route::get('/articles/edit/{art_id}', 'ArticlesController@edit');

回答4:

I recommend to work with named routes!

Your routes/web.php file:

Route::get('articles/{articleId}/edit', 'YourController@action')->name('article.edit');

Your Blade-Template file:

<a href=" {{ route('article.edit', ['articleId' => $article->id]) }}">Edit></a>