php 如何在 Laravel 5 中使用分页？
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How to use pagination in laravel 5?
提问by Sojan V Jose
am trying to port my laravel4 application to laravel 5 . In the previous version i could use the following method for generating pagination urls .
我正在尝试将我的 laravel4 应用程序移植到 laravel 5。在以前的版本中，我可以使用以下方法生成分页网址。
In controller:
在控制器中：
$this->data['pages']= Page::whereIn('area_id', $suburbs)->where('score','>','0')->orderBy('score','desc')->paginate(12);
and after sharing the data array with the view, i could user
在与视图共享数据数组后，我可以使用
In views:
在视图中：
{{$pages->links()}}
In laravel 5 the doing that results in following error
在 Laravel 5 中，这样做会导致以下错误
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 ?
不确定我在这里缺少什么，有人可以帮忙吗？
回答by Alasu Paul Sabrin
In Laravel 5 there is no method "links" you can try this
在 Laravel 5 中没有方法“链接”你可以试试这个
{!! $pages->render() !!}
回答by David
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:
在其他框架中，分页可能非常痛苦。Laravel 5 让它变得轻而易举。要使用它，首先您必须更改从数据库调用数据的控制器代码：
public function index()
{
$users = DB::table('books')->simplePaginate(5);
//using pagination method
return view('index', ['users' => $users]);
}
...after that you can use this code:
...之后，您可以使用此代码：
<?php echo $users->render(); ?>
That will make you use simple Laravel 5 beauty.
这将使您使用简单的 Laravel 5 美。
回答by Nadeem Qasmi
Pagination Laravel 5.6.26, for the pagination the controller are :
分页 Laravel 5.6.26，对于分页，控制器是：
Controller code (https://laravel.com/docs/5.6/pagination#basic-usage)
控制器代码（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)
前端进入刀片（视图）（https://laravel.com/docs/5.6/pagination#displaying-pagination-results）
{{ $users->links() }}
回答by Jose Mhlanga
$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 {{ $items->links() }}
</div>
Use the array name (items) in the render() or links() method NOT the array item. It worked for me.
在 render() 或 links() 方法中使用数组名称（项目）而不是数组项目。它对我有用。
回答by Rahul Saha
@if ($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"><</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">></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
回答by diego
Hi there is my code for pagination:
Use in view:
@include('pagination.default', ['paginator' => $users])
嗨，我的分页代码：在视图中使用：
@include('pagination.default', ['paginator' => $users])
Views/pagination/default.blade.php
视图/分页/default.blade.php
@if ($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
