How to set pagination in laravel 5 while passing an array from the controller

Question

Here is my code

Controller

$query = "SELECT `h`.*,".$countquery.",".$minquery." FROM `abserve_hotels` as `h`";
$aReslts = $this->model->getData($query,$sqlCond);

$this->data['Rooms'] =!empty($aReslts) ? $aReslts : '';

return view('hotel.list', $this->data);

Here,$sqlCond returns several result set while passing Ajax and for post action

View

<?php
if(($Rooms)){
?>
@foreach($Rooms as $room)
{{ $room->id }}
@endforeach

<?php
}
?>

I tried several ways to set the pagination methods..But it din't work for me..

How should I make pagination for this..Could someone help me!!..

Thanks in advance

brainless · Accepted Answer · 2016-03-11 13:10:10Z

2

You can create a paginator manually by using

$pagination = Paginator::make($this->data['Rooms'], count($this->data['Rooms']), 5);

To display links use echo $pagination->links(); Or {{ $pagination->links() }}

answered Mar 11, 2016 at 13:10

brainless

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Comments

Arash Hatami · Accepted Answer · 2019-07-14 14:09:10Z

2

Use this way .... very easy

public function items(Request $request)
{
    $items = [
        'item1',
        'item2',
        'item3',
        'item4',
        'item5',
        'item6',
        'item7',
        'item8',
        'item9',
        'item10'
        ];

    // Get current page form url e.x. &page=1
    $currentPage = LengthAwarePaginator::resolveCurrentPage();

    // Create a new Laravel collection from the array data
    $itemCollection = collect($items);

    // Define how many items we want to be visible in each page
    $perPage = 1;

    // Slice the collection to get the items to display in current page
    $currentPageItems = $itemCollection->slice(($currentPage * $perPage) - $perPage, $perPage)->all();

    // Create our paginator and pass it to the view
    $paginatedItems= new LengthAwarePaginator($currentPageItems , count($itemCollection), $perPage);

    // set url path for generated links
    $paginatedItems->setPath($request->url());

    return view('items_view', ['items' => $paginatedItems]);
}

answered Jul 14, 2019 at 14:09

Arash Hatami

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1 Comment

Albert George Over a year ago

Thanks. It worked for me!! Also in blade file I added {{ $paginatedItems->links() }} to display pagination Links.

Shyam Achuthan · Accepted Answer · 2016-03-11 07:34:58Z

0

public function controllerFunction($page=1)
{
   $limit = 10;
   $start = ($page-1)*$limit;
   $query = "SELECT `h`.*,".$countquery.",".$minquery." FROM `abserve_hotels` as `h`";
$aReslts = $this->model->getData($query,$sqlCond)->skip($start)->take($limit);

$this->data['Rooms'] =!empty($aReslts) ? $aReslts : '';

return view('hotel.list', $this->data);
}

This will give you paginated content. You can add pagination on your view via laravel pagination or using a custom code.

answered Mar 11, 2016 at 7:34

Shyam Achuthan

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3 Comments

Suganya Rajasekar Over a year ago

Using this code in controller and adding ` echo $aHotelRooms->render(); ` in view may set pagination!!...

Suganya Rajasekar Over a year ago

And in my controller function I already pass a Request $request how should I give the $page=1 by using , !!...

Suganya Rajasekar Over a year ago

When use echo $Rooms->render() I'm always getting 1 FatalErrorException in 3d4c6f287876462533e51d881f81767c line 45: Call to a member function render() on array what should I do for it.. : (

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