How to use pagination in laravel 5?

Question

am trying to port my laravel4 application to laravel 5 . In the previous version i could use the following method for generating pagination urls .

In controller:

$this->data['pages']= Page::whereIn('area_id', $suburbs)->where('score','>','0')->orderBy('score','desc')->paginate(12);

and after sharing the data array with the view, i could user

In views :

{{$pages->links()}}

In laravel 5 the doing that results in following error

ErrorException in AbstractPaginator.php line 445:
call_user_func_array() expects parameter 1 to be a valid callback, class 'Illuminate\Support\Collection' does not have a method 'links'

not sure what i am missing here, can somebody help ?

I suggest to wait until Laravel 5 is stable because it has some bugs and the base code can change. — Ludo237
– Ludo237, Commented Jan 16, 2015 at 9:52
@ClaudioLudovicoPanetta i think it will come out this month :) . just wanted to be in sync with the changes . using 4.2 for production though . — Sojan Jose
– Sojan Jose, Commented Jan 16, 2015 at 9:54
your problem seems more a Laravel issue than yours, I will try to replicate you problem for a solution, brb. — Ludo237
– Ludo237, Commented Jan 16, 2015 at 10:02
@ClaudioLudovicoPanetta thanks :) . will wait and see. else will add a issue at github — Sojan Jose
– Sojan Jose, Commented Jan 16, 2015 at 10:06

Limon Monte · Accepted Answer · 2015-04-19 10:15:05Z

73

In Laravel 5 there is no method "links" you can try this

{!! $pages->render() !!}

edited Apr 19, 2015 at 10:15

Limon Monte

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answered Jan 16, 2015 at 15:39

Alasu Paul Sabrin

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4 Comments

Sojan Jose Over a year ago

{!!$pages->render()!!} did the trick , the other syntax , in laravel5 escapes html . thanks

hfingler Over a year ago

Finally a solution to a simple pagination, not this new "create 5 php files, implement a class and write 200 lines of code". FINALLY! Thank you.

Muhaimin Over a year ago

this doesn't work. Is there anything that I need to prepare first before using it?

Vijay Rana Over a year ago

@muhaimin You need to attach paginate() function while accessing data from database. for example: Page::whereIn('area_id', $suburbs)->orderBy('score','desc')->paginate(12); and pass the variable to the view.

sideshowbarker · Accepted Answer · 2015-09-22 11:34:34Z

9

In other frameworks, pagination can be very painful. Laravel 5 makes it a breeze. For using it, first you have to make a change in your controller code where you calling data from the database:

 public function index()
{       
        $users = DB::table('books')->simplePaginate(5);
        //using pagination method
        return view('index', ['users' => $users]);
}

...after that you can use this code:

<?php echo $users->render(); ?>

That will make you use simple Laravel 5 beauty.

edited Sep 22, 2015 at 11:34

sideshowbarker

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answered Sep 22, 2015 at 11:13

David

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Comments

VIKAS KATARIYA · Accepted Answer · 2024-06-03 13:48:25Z

6

Pagination Laravel 5.6.26, for the pagination the controller are :

Controller code (https://laravel.com/docs/5.6/pagination#basic-usage)

    $posts = Post::orderBy('created_at','desc')->paginate(10);
    return view('posts.index')->with('posts', $posts);

Front end into blade (view) (https://laravel.com/docs/5.6/pagination#displaying-pagination-results)

   {{ $users->links() }}

edited Jun 3, 2024 at 13:48

VIKAS KATARIYA

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answered Jul 3, 2018 at 13:18

Nadeem Qasmi

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Comments

Jose Mhlanga · Accepted Answer · 2019-06-12 12:32:37Z

5

$items = SomeDataModel->get()->paginate(2); // in your controller

@foreach(items as $item)   // in your view.blade file
....echo some data here
@endforeach

<div class="pagination">
    {{ $items->render() }} or {{ $items->links() }}
</div>

Use the array name (items) in the render() or links() method NOT the array item. It worked for me.

edited Jun 12, 2019 at 12:32

answered Jun 12, 2019 at 12:27

Jose Mhlanga

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Comments

Raj Kanani · Accepted Answer · 2022-02-12 09:28:50Z

2

Best Way Pagination And Simple Code

 $limit = 10;
 $page_no = ($request->input('**page_no**') * $limit) - $limit;


   $data = User::whereraw($whereraw)
                  ->limit(**$limit**)
                  ->offset(**$page_no**)
                  ->get();

edited Feb 12, 2022 at 9:28

answered Sep 4, 2021 at 10:40

Raj Kanani

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Comments

OmidDarvishi · Accepted Answer · 2022-05-01 07:47:51Z

1

use

$users = User::where('id', Auth::user()->id)->simplePaginate(20);

and use in blade

{!! $users->render() !!}

edited May 1, 2022 at 7:47

answered Jul 5, 2020 at 12:49

OmidDarvishi

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1 Comment

Ahmed Numaan Over a year ago

Undefined variable: pages

Rahul Saha · Accepted Answer · 2016-12-14 11:22:58Z

@if ($posts->lastPage() > 1)
        <nav aria-label="Page navigation">
            <ul class="pagination">
                @if($posts->currentPage() != 1 && $posts->lastPage() >= 5)
                <li>
                    <a href="{{ $posts->url($posts->url(1)) }}" aria-label="Previous">
                        <span aria-hidden="true">First</span>
                    </a>
                </li>
                @endif
                @if($posts->currentPage() != 1)
                <li>
                    <a href="{{ $posts->url($posts->currentPage()-1) }}" aria-label="Previous">
                        <span aria-hidden="true">&#x3C;</span>
                    </a>
                </li>
                @endif
                @for($i = max($posts->currentPage()-2, 1); $i <= min(max($posts->currentPage()-2, 1)+4,$posts->lastPage()); $i++)
                @if($posts->currentPage() == $i)
                <li class="active">
                @else
                <li>
                @endif
                    <a href="{{ $posts->url($i) }}">{{ $i }}</a>
                </li>
                @endfor
                @if ($posts->currentPage() != $posts->lastPage())
                <li>
                    <a href="{{ $posts->url($posts->currentPage()+1) }}" aria-label="Next">
                        <span aria-hidden="true">&#x3E;</span>
                    </a>
                </li>
                @endif
                @if ($posts->currentPage() != $posts->lastPage() && $posts->lastPage() >= 5)
                <li>
                    <a href="{{ $posts->url($posts->lastPage()) }}" aria-label="Next">
                        <span aria-hidden="true">Last</span>
                    </a>
                </li>
                @endif
            </ul>
        </nav>
        @endif

Give some explanation, the answer should be explained atleast a bit.

diego · Accepted Answer · 2015-06-12 00:52:39Z

Hi there is my code for pagination: Use in view:
@include('pagination.default', ['paginator' => $users])

Views/pagination/default.blade.php

@if ($paginator->lastPage() > 1)
    <ul class="pagination">
        <!-- si la pagina actual es distinto a 1 y hay mas de 5 hojas muestro el boton de 1era hoja -->
        <!-- if actual page is not equals 1, and there is more than 5 pages then I show first page button -->
        @if ($paginator->currentPage() != 1 && $paginator->lastPage() >= 5)
            <li>
                <a href="{{ $paginator->url($paginator->url(1)) }}" >
                    <<
                </a>
            </li>
        @endif

        <!-- si la pagina actual es distinto a 1 muestra el boton de atras -->
        @if($paginator->currentPage() != 1)
            <li>
                <a href="{{ $paginator->url($paginator->currentPage()-1) }}" >
                    <
                </a>
            </li>
        @endif

        <!-- dibuja las hojas... Tomando un rango de 5 hojas, siempre que puede muestra 2 hojas hacia atras y 2 hacia adelante -->
        <!-- I draw the pages... I show 2 pages back and 2 pages forward -->
        @for($i = max($paginator->currentPage()-2, 1); $i <= min(max($paginator->currentPage()-2, 1)+4,$paginator->lastPage()); $i++)
                <li class="{{ ($paginator->currentPage() == $i) ? ' active' : '' }}">
                    <a href="{{ $paginator->url($i) }}">{{ $i }}</a>
                </li>
        @endfor

        <!-- si la pagina actual es distinto a la ultima muestra el boton de adelante -->
        <!-- if actual page is not equal last page then I show the forward button-->
        @if ($paginator->currentPage() != $paginator->lastPage())
            <li>
                <a href="{{ $paginator->url($paginator->currentPage()+1) }}" >
                    >
                </a>
            </li>
        @endif

        <!-- si la pagina actual es distinto a la ultima y hay mas de 5 hojas muestra el boton de ultima hoja -->
        <!-- if actual page is not equal last page, and there is more than 5 pages then I show last page button -->
        @if ($paginator->currentPage() != $paginator->lastPage() && $paginator->lastPage() >= 5)
            <li>
                <a href="{{ $paginator->url($paginator->lastPage()) }}" >
                    >>
                </a>
            </li>
        @endif
    </ul>
@endif

look that i didn't copy well... the code starts before the gray area.

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