Laravel's paging is very simple and easy to use, but there is one thing that is not very ideal.
The link generated by the default paging is
users?page=1
This form is not very friendly to SEO (theory) Above)
Then we may need a paging link in the form of
users/page/1
So how to do it?
I now have two ways to implement it
The first one was used in my previous project, directly regularizing
Directly regularizing in blade There should be room for optimization in replacing
{!! preg_replace("~(/?)\?page=~", '/page/', $users->render()) !!}
. I hope you can give me some advice.
Of course, the corresponding routing must also be configured accordingly. It is very simple, so I won’t expand it.
The effect of this method is achieved, but it really doesn’t have the style of laravel.
What is the style of laravel? composer, package
So let’s introduce the second method
laravel-paginateroute
The method of use is very simple, the following is excerpted from the readme
The first step
// 要注意对应版本,可到https://packagist.org 方便查看 composer require spatie/laravel-paginateroute
The second step
Register,
// config/app.php 'providers' => [ ... 'Spatie\PaginateRoute\PaginateRouteServiceProvider', ]; 'aliases' => [ ... 'PaginateRoute' => 'Spatie\PaginateRoute\PaginateRouteFacade', ];
The third step
// app/Providers/RouteServiceProvider.php use PaginateRoute; // ... public function boot(Router $router) { PaginateRoute::registerMacros(); parent::boot($router); }
Use
In routing
// app/Http/routes.php // Generates /users & /users/page/{page} Route::paginate('users', 'UsersController@index');
This will generate two routes
/users /users/page/{page}
Then in the controller, you can write it normally according to laravel's paging
Used like this in the page, this is the corresponding writing method of simplePaginate
@if(PaginateRoute::hasPreviousPage()) <a href="{{ PaginateRoute::previousPageUrl() }}">Previous</a> @endif @if(PaginateRoute::hasNextPage($users)) <a href="{{ PaginateRoute::nextPageUrl($users) }}">Next</a> @endif
Recommended: "laravel tutorial"
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