Laravel 5.7 Pagination Pretty URL

Question

I want to create a pagination for post controller, this is my route:

Route::get('post/{page?}' , 'PostController@Test');

And my action:

public function Test($page = 1)
{
    $tests = Test::paginate(15, ['*'], 'page', $page);

    foreach ($tests as $test) {
        echo $test->id;
    }

    echo $tests->render();
}

Now I can go to http://127.0.0.1:8000/post/2 to open page 2, but my pagination url looks like http://127.0.0.1:8000/post?page=2

Can i fix this?

I think [this](https://stackoverflow.com/questions/20974404/laravel-pagination-pretty-url) might help. — Ishaan, Jan 27 '19 at 09:08
BTW, when you go to `http://127.0.0.1:8000/post/2`, it will give you the second post in your db. pagination is different. — GabMic, Jan 27 '19 at 09:08
I want `echo $tests->render();` print urls like `http://127.0.0.1:8000/post/2` `http://127.0.0.1:8000/post/3` @GabMic — Masoud92m, Jan 27 '19 at 09:14
I want `http://127.0.0.1:8000/post/2` give me second page, i get posts with slug @GabMic — Masoud92m, Jan 27 '19 at 09:16

score 0 · Answer 1 · answered Jan 27 '19 at 09:12

As far as I know, it is not possible to introduce pretty URL for Pagination in Laravel without using external Package.

In one of my project, I was able to build the same using Laravel Paginateroute Package.

which you can install by this command :

composer require spatie/laravel-paginateroute

I have followed the Documentation provided by the author to implement the same.

Laravel 5.7 Pagination Pretty URL

1 Answers1