Laravel Pagination links not including other GET parameters

EDIT: Connor's comment with Mehdi's answer are required to make this work. Thanks to both for their clarifications.

->appends() can accept an array as a parameter, you could pass Input::except('page'), that should do the trick.

Example:

return view('manage/users', [
    'users' => $users->appends(Input::except('page'))
]);

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I think you should use this code in Laravel version 5+. Also this will work not only with parameter page but also with any other parameter(s):

$users->appends(request()->input())->links();

Personally, I try to avoid using Facades as much as I can. Using global helper functions is less code and much elegant.

UPDATE:

Do not use Input Facade as it is deprecated in Laravel v6+

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You could use

->appends(request()->query())

Example in the Controller:

$users = User::search()->order()->with('type:id,name')
    ->paginate(30)
    ->appends(request()->query());

return view('users.index', compact('users'));

Example in the View:

{{ $users->appends(request()->query())->links() }}

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Be aware of the Input::all() , it will Include the previous ?page= values again and again in each page you open !
for example if you are in ?page=1 and you open the next page, it will open ?page=1&page=2
So the last value page takes will be the page you see ! not the page you want to see

Solution : use Input::except(array('page'))

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Laravel 7.x and above has added new method to paginator:

->withQueryString()

So you can use it like:

{{ $users->withQueryString()->links() }}

For laravel below 7.x use:

{{ $users->appends(request()->query())->links() }}

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Not append() but appends() So, right answer is:

{!! $records->appends(Input::except('page'))->links() !!}

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LARAVEL 5

The view must contain something like:

{!! $myItems->appends(Input::except('page'))->render() !!}

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Use this construction, to keep all input params but page

{!! $myItems->appends(Request::capture()->except('page'))->render() !!}

Why?

1) you strip down everything that added to request like that

  $request->request->add(['variable' => 123]);

2) you don't need $request as input parameter for the function

3) you are excluding "page"

PS) and it works for Laravel 5.1

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Include This In Your View Page

 $users->appends(Input::except('page'))

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for who one in laravel 5 or greater in blade:

{{ $table->appends(['id' => $something ])->links() }}

you can get the passed item with

$passed_item=$request->id;

test it with

dd($passed_item);

you must get $something value

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In Laravel 7.x you can use it like this:

{{ $results->withQueryString()->links() }}

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Pass the page number for pagination as well. Some thing like this

$currentPg = Input::get('page') ? Input::get('page') : '1'; $boards = Cache::remember('boards'.$currentPg, 60, function(){ return WhatEverModel::paginate(15); });

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In Your controller after pagination add withQueryString() like below

$post = Post::paginate(10)->withQueryString();

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Many solution here mention using Input...

Input has been removed in Laravel 6, 7, 8

Use Request instead.

Here's the blade statement that worked in my Laravel 8 project:

{{$data->appends(Request::except('page'))->links()}}

Where $data is the PHP object containing the paginated data.

Thanks to Alexandre Danault who pointed this out in this comment.

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