How to make perfect URL using paginate in Laravel?

My Controller

$posts = Posts::where('status','publish')->orderBy('id', 'desc')->take(10)->get();
return view('front', compact('posts'));

My View

{{$posts->links}}

I want convert

http://127.0.0.1:8000/?page=3

http://127.0.0.1:8000/page/3

Thanks

jlrdw

5 years ago

Level 75

You have to do some custom coding for that. You could use the length aware pagination class.

https://laravel.com/api/8.x/Illuminate/Pagination/LengthAwarePaginator.html

It's easier to have pretty url's except for the ?page= part. I would leave as is.

You could download https://github.com/spatie/laravel-paginateroute and study how they did it.

But

That package is no longer maintained, just use as a guide to learn how to custom code it yourself.

sauravs012

5 years ago

Level 2

I've tried custom code, but not works

jlrdw

5 years ago

Level 75

Did you study the code used in the spatie library to get some ideas.

MichalOravec

5 years ago

Level 75

You can use this package

https://github.com/michaloravec/laravel-paginateroute

There it can be like

/news/page/2

// or 

/news/page-2

// or

/news/2

1 like

Čamo

4 years ago

Level 3

How do you render the html?

MichalOravec

4 years ago

Level 75

Do you mean pagination links? In the same way like in Laravel, with custom pagination view.

https://laravel.com/docs/8.x/pagination#customizing-the-pagination-view

{{ $items->links('custom.pagination') }}

For example views/custom/pagination.blade.php

@if ($paginator->hasPages())
    <nav class="text-center">
        <ul class="pagination pagination-md">
            {{-- Previous Page Link --}}
            @if ($paginator->onFirstPage())
                <li class="disabled"><span>@lang('pagination.previous')</span></li>
            @else
                <li><a href="{{ PaginateRoute::previousPageUrl() }}" rel="prev">@lang('pagination.previous')</a></li>
            @endif

            {{-- Pagination Elements --}}
            @foreach ($elements as $element)
                {{-- "Three Dots" Separator --}}
                @if (is_string($element))
                    <li class="disabled"><span>{{ $element }}</span></li>
                @endif

                {{-- Array Of Links --}}
                @if (is_array($element))
                    @foreach ($element as $page => $url)
                        @if ($page == $paginator->currentPage())
                            <li class="active"><span>{{ $page }}</span></li>
                        @else
                            @if (in_array($paginator->currentPage(), [$page - 1, $page, $page + 1]))
                                <li><a href="{{ PaginateRoute::pageUrl($page) }}">{{ $page }}</a></li>
                            @else
                                <li class="hidden-xs"><a href="{{ PaginateRoute::pageUrl($page) }}">{{ $page }}</a></li>
                            @endif
                        @endif
                    @endforeach
                @endif
            @endforeach

            {{-- Next Page Link --}}
            @if ($paginator->hasMorePages())
                <li><a href="{{ PaginateRoute::nextPageUrl($paginator) }}" rel="next">@lang('pagination.next')</a></li>
            @else
                <li class="disabled"><span>@lang('pagination.next')</span></li>
            @endif
        </ul>
    </nav>
@endif