How to remove query string page from the first page of laravel pagination?

Question

I'm using laravel-5.4 pagination like the following:

public function index()
    {
        $jobs = Job::paginate(5);
        return view('job.index', compact('jobs'));
    }

In the view:

{{ $jobs->links() }}

There is a problem of generating two typical pages: /job and /job?page=1 the two page has the same contents.

I want to do anything that removes the query string page from the first page of the pagination.

I have tried the following:

if ($jobs->onFirstPage()){
  $jobs->setPageName('');
}

But this corrupt the pagination, i.e the links of pages does not load correctly and the query string value remains for all pages.

SaidbakR · Accepted Answer · 2018-03-09T01:08:26.887

The effective solution for this issue that I have found is to edit the pagination template.

First publish the pagination template from the vendors using the following command from the root of the project:

php artisan vendor:publish --tag=laravel-pagination

Now a file at resources/views/vendor/pagination/default.blade.php should be found and it could be edited like the following using str_replace() for the urls of each page and back navigation button:

<li><a href="{{str_replace('?page=1','',$paginator->previousPageUrl())}}" title="{{__('Previous')}}" rel="prev">{{$foxPrev}}</a></li>

and

 <li><a href="{{str_replace('?page=1','',$url)}}">{{ $page }}</a></li>

Update:

A bug was found with ?page=10 so instead of using str_replace we should using preg_replace like the following:

<li><a href="{{preg_replace('/\?page=[1]$/','',$url)}}">{{ $page }}</a></li>

Update 2:

In case of using any customized name for the page number parameter other than page, we could use the paginator getter for the $pageName property like the following in the pagination template:

<li><a href="{{preg_replace('/\?'.$paginator->getPageName().'=[1]$/','',$url)}}">{{ $page }}</a></li>

To know more about how to use more than one pagination on the same page or how to customize the page number parameter $pageName from view, checkout this answer

Expression '/((\?|\&)page=[1]$)|(page=[1]($|\&))/' OR '/((\?|\&)'.$paginator->getPageName().'=[1]$)|('.$paginator->getPageName().'=[1]($|\&))/' will cover all the cases. For example: '...?page=1', '...?page=1&brand=2', '...?brand=2&page=1&filter=3', '...?brand=2&page=1'. And it will not react on 'page=10' in any position. — Max Basmanov, Oct 15 '22 at 03:11
I don't understand how this question is upvoted when removing all parameters from link will give us URL to the same page with same parameters. Like if we are on the page ?page=2 it will give us same link on Page 1. — KorbenDallas, Nov 09 '22 at 19:03

score 4 · Answer 2 · answered Aug 28 '18 at 14:36

You can extend LengthAwarePaginator method url($page)

/**
 * Get the URL for a given page number.
 *
 * @param  int  $page
 * @return string
 */
public function url($page)
{
    if ($page <= 0) {
        $page = 1;
    }

    // If we have any extra query string key / value pairs that need to be added
    // onto the URL, we will put them in query string form and then attach it
    // to the URL. This allows for extra information like sortings storage.
    $parameters = ($page > 1) ? [$this->pageName => $page] : [];

    if (count($this->query) > 0) {
        $parameters = array_merge($this->query, $parameters);
    }

    return rtrim($this->path
        .(Str::contains($this->path, '?') ? '&' : '?')
        .http_build_query($parameters, '', '&')
        .$this->buildFragment(), '?');
}

This should be the accepted answer. Remove `['page' => 1]` from the `$parameters` array, and then remove the trailing `?` if necessary. EZPZ. Thanks. :) — keyboardSmasher, Feb 18 '19 at 16:37

score 3 · Answer 3 · answered Oct 20 '20 at 09:38

improved Aksi answer

app/Services/CustomLengthAwarePaginator.php

<?php

namespace App\Services;

use Illuminate\Pagination\LengthAwarePaginator;
use Illuminate\Support\Arr;
use Illuminate\Support\Str;

class CustomLengthAwarePaginator extends LengthAwarePaginator
{
    /**
     * Get the URL for a given page number.
     *
     * @param  int  $page
     * @return string
     */
    public function url($page)
    {
        if ($page <= 0) {
            $page = 1;
        }

        // If we have any extra query string key / value pairs that need to be added
        // onto the URL, we will put them in query string form and then attach it
        // to the URL. This allows for extra information like sortings storage.
        $parameters = ($page > 1) ? [$this->pageName => $page] : [];

        if (count($this->query) > 0) {
            $parameters = array_merge($this->query, $parameters);
        }

        return $this->path()
            . (count($parameters) > 0
                ? (Str::contains($this->path(), '?') ? '&' : '?')
                : '')
            . Arr::query($parameters)
            . $this->buildFragment();
    }
}

in AppServiceProvider.php or another

public function boot()
{
    app()->bind(LengthAwarePaginator::class, CustomLengthAwarePaginator::class);
}

kai ogita · Answer 4 · 2017-03-14T01:46:23.377

0

add canonical tag in if you care SEO

i don't have any idea...

if($pageNum==1){
    return redirect()->to($path);
}

$pageNum id get Request or urself Class

$path = "/jobs"

edited Mar 14 '17 at 01:46

answered Mar 14 '17 at 01:25

kai ogita

136
2
9

1

This for SEO, so what's about extra resources and traffic consumption to retrieve the same content?! Unfortunately this is not the ideal solution that I have to find. – SaidbakR Mar 14 '17 at 01:28

score 0 · Answer 5 · answered Apr 18 '17 at 09:45

In your routes, give a name to the route which leads to your function:

Route::get('/yourRoute','YourController@foo')->name('yourRouteName');

Then, In your function in the controller, use this:

public function foo() {
    if( request()->page=='1' )
    {
            return redirect()->route('yourRouteName',[$id]);
    }
     else {
            // Your function content
    }
}

score 0 · Answer 6 · answered Apr 18 '17 at 10:05

0

I think you can try this:

{{ $jobs->links() }}

To

{{ $jobs->nextPageUrl() }}

Hope this work for you!

answered Apr 18 '17 at 10:05

AddWeb Solution Pvt Ltd

21,025
5
26
57

score 0 · Answer 7 · answered Mar 20 '18 at 15:59

0

You can simply modify the view like this:

@if ($page==1)
   <a href="{{ str_replace('?page=1', '', $url) }}">{{ $page }}</a>
@else
   <a href="{{ $url }}">{{ $page }}</a>
@endif

answered Mar 20 '18 at 15:59

Dong3000

566
2
7
24

Eric · Answer 8 · 2018-10-06T21:51:03.000

I found the following works and takes into account page=10 issues:

After you export your pagination views, make the following 2 modifications:

MODIFICATION 1

Replace this:

{{ $paginator->previousPageUrl() }}

with this:

{{ preg_replace('/(?:(&|\\?)page=[1])(?!\\d)/ui','', $paginator->previousPageUrl()) }}

MODIFICATION 2

Replace this:

{{ $url }}

with this:

{{ preg_replace('/(?:(&|\\?)page=[1])(?!\\d)/ui','', $url ) }}

score 0 · Answer 9 · answered Apr 07 '19 at 10:17

0

I've done it this way:

{{ $posts->currentPage() == 2 ? route('home') : $posts->previousPageUrl() }}

answered Apr 07 '19 at 10:17

wast

878
9
27

score 0 · Answer 10 · answered Mar 08 '21 at 23:10

Just create a Middleware by using this command

php artisan make:middleware RemovePageQuery

Write this code inside middleware

<?php

namespace App\Http\Middleware;

use Closure;
use Illuminate\Http\Request;

class RemovePageQuery
{
    /**
     * Handle an incoming request.
     *
     * @param  \Illuminate\Http\Request  $request
     * @param  \Closure  $next
     * @return mixed
     */
    public function handle(Request $request, Closure $next)
    {
        if ($request->page == 1)
        {
            return redirect(url()->current());
        }
        
        return $next($request);
    }
}

Add middleware global in App/Http/Kernel.php inside $middleware.

 protected $middleware = [
        // \App\Http\Middleware\TrustHosts::class,
        \App\Http\Middleware\RemovePageQuery::class, //Here
        \App\Http\Middleware\TrustProxies::class,
        \Fruitcake\Cors\HandleCors::class,
        \App\Http\Middleware\PreventRequestsDuringMaintenance::class,
        \Illuminate\Foundation\Http\Middleware\ValidatePostSize::class,
        \App\Http\Middleware\TrimStrings::class,
        \Illuminate\Foundation\Http\Middleware\ConvertEmptyStringsToNull::class,
    ];

score 0 · Answer 11 · answered Apr 16 '21 at 08:37

I was searching to fix similar issue to prevent content duplication but for symfony not laravel. I fixed it in my way and decided to publish it. May be it will help somebody.

In symfony I use KNP paginator. It provides paginator template, like this:

     <a href="{{ path(route, query|merge({(pageParameterName): page})) }}">{{ page }}</a>

'Query' array contains route params and the task is to remove page param if it == 1.

Very simple:

    {% set page_seo = (page == 1) ? null : page %}
    <a href="{{ path(route, query|merge({(pageParameterName): page_seo})) }}">{{ page }}</a>

Actually - set to null page if it == 1.

score 0 · Answer 12 · answered Oct 27 '22 at 07:57

0

By this way you can remove form field from the request in Laravel Controller. It will work 100%. You will get in request without page param.

   $request->request->remove('page');

answered Oct 27 '22 at 07:57

Kaleemullah

446
3
8

KorbenDallas · Answer 13 · 2022-11-10T20:50:53.310

Better answer than @SaidbakR has provided, as I it's not right. Why? Because you will receive link for the same page as URL will be empty. And for example if you make request from url?page=2, you'll get link to same url?page=2 where link to first page must appear.

My easy variant:

Also publish pagination views: php artisan vendor:publish --tag=laravel-pagination
Go to /views/vendor/pagination/default.blade.php

Add elseif case to previous page link:

     @elseif($paginator->currentPage() === 2)
         <li class="pagination-next-prev link">
             <a class="prev page-url" href="?{{ http_build_query(Arr::except(Request::query(), 'page')) }}" rel="prev"
                aria-label="@lang('pagination.previous')">
                 Prev
             </a>
         </li>
     @else

Add similar elseif to array of links:

@elseif(1 === $page)
   <li class="page">
       <a class="page-url" href="?{{ http_build_query(Arr::except(Request::query(), 'page')) }}">
           {{ $page }}
       </a>
   </li>
 @else

This case is also working if you are using ajax request to another page for pagination.
Done

You will have link to the page you opened, but without params. Yes it will be showed as ? in the URL, but it's fair price to avoid duplications of URLs, as link with just question mark for google same as without.

score 0 · Answer 14 · answered Aug 16 '23 at 15:28

run php artisan vendor:publish --tag=laravel-pagination

update resources/views/vendor/pagination/bootstrap-5.blade.php to this

@if ($paginator->hasPages())
    <nav class="d-flex justify-items-center justify-content-between">
        <div class="d-flex justify-content-between flex-fill d-sm-none">
            <ul class="pagination">
                {{-- Previous Page Link --}}
                @if ($paginator->onFirstPage())
                    <li class="page-item disabled" aria-disabled="true">
                        <span class="page-link">@lang('pagination.previous')</span>
                    </li>
                @else
                    <li class="page-item">
                        <a class="page-link" href="{{ preg_replace('/[\&\?]page=[1]$/','',$paginator->previousPageUrl()) }}" rel="prev">@lang('pagination.previous')</a>
                    </li>
                @endif

                {{-- Next Page Link --}}
                @if ($paginator->hasMorePages())
                    <li class="page-item">
                        <a class="page-link" href="{{ $paginator->nextPageUrl() }}" rel="next">@lang('pagination.next')</a>
                    </li>
                @else
                    <li class="page-item disabled" aria-disabled="true">
                        <span class="page-link">@lang('pagination.next')</span>
                    </li>
                @endif
            </ul>
        </div>

        <div class="d-none flex-sm-fill d-sm-flex align-items-sm-center justify-content-sm-between">


            <div>
                <ul class="pagination">
                    {{-- Previous Page Link --}}
                    @if ($paginator->onFirstPage())
                        <li class="page-item disabled" aria-disabled="true" aria-label="@lang('pagination.previous')">
                            <span class="page-link" aria-hidden="true">&lsaquo;</span>
                        </li>
                    @else
                        <li class="page-item">
                            <a class="page-link" href="{{ preg_replace('/[\&\?]page=[1]$/','',$paginator->previousPageUrl()) }}" rel="prev" aria-label="@lang('pagination.previous')">&lsaquo;</a>
                        </li>
                    @endif

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

                        {{-- Array Of Links --}}
                        @if (is_array($element))
                            @foreach ($element as $page => $url)
                                @if ($page == $paginator->currentPage())
                                    <li class="page-item active" aria-current="page"><span class="page-link">{{ $page }}</span></li>
                                @else
                                    <li class="page-item"><a class="page-link" href="{{ preg_replace('/[\&\?]page=[1]$/','',$url) }}">{{ $page }}</a></li>
                                @endif
                            @endforeach
                        @endif
                    @endforeach

                    {{-- Next Page Link --}}
                    @if ($paginator->hasMorePages())
                        <li class="page-item">
                            <a class="page-link" href="{{ $paginator->nextPageUrl() }}" rel="next" aria-label="@lang('pagination.next')">&rsaquo;</a>
                        </li>
                    @else
                        <li class="page-item disabled" aria-disabled="true" aria-label="@lang('pagination.next')">
                            <span class="page-link" aria-hidden="true">&rsaquo;</span>
                        </li>
                    @endif
                </ul>
            </div>

            <div>
                <p class="small text-muted">
                    {!! __('Showing') !!}
                    <span class="fw-semibold">{{ $paginator->firstItem() }}</span>
                    {!! __('to') !!}
                    <span class="fw-semibold">{{ $paginator->lastItem() }}</span>
                    {!! __('of') !!}
                    <span class="fw-semibold">{{ $paginator->total() }}</span>
                    {!! __('results') !!}
                </p>
            </div>
        </div>
    </nav>
@endif

How to remove query string page from the first page of laravel pagination?

14 Answers14

Update:

Update 2: