Question

我试图实现像sphinx文档这样的视图，你可以点击next和previous来在页面之间导航。不确定我是否在阵列右侧使用自定义分页器，我无法获得页面导航链接以显示在视图页面上。这是代码：

   public function paginate($array, $perPage, $pageStart=1) {

    $offset = ($pageStart * $perPage) - $perPage;

    return new Paginator(array_slice($array, $offset, $perPage, true), $perPage, $pageStart);
   }

   view()->composer('layouts.book', function($view)
   {
        //some other code
        $pages = [];
        foreach($book->textsection_pages as $textsection) {
            $pages[] = $textsection->alias;
        }
        foreach($book->task_pages as $task) {
            $pages[] = $task->alias;
        }
        foreach($book->tutor_pages as $tutor) {
            $pages[] = $tutor->alias;
        }
        foreach($book->eval_pages as $eval) {
            $pages[] = $eval->alias;
        }

        $chapters = $book->chapters()->orderBy('weight', 'asc')->get();
        $paginated = $this->paginate($pages, 1);
        $view->with(['chapters' => $chapters, 'book' => $book, 'paginated' => $paginated]);

    });

在视图中我调用了{!! $paginated->render() !!}，但没有显示导航链接。

Answer 1

以下是我在应用程序中处理自定义分页的方法（Laravel 5.1）。我创建了部分视图resources/views/partials/pagination.blade.php

@if ($paginator->lastPage() > 1)
<ul id="pagination">
    <li>
    @if ($paginator->currentPage() > 1)
        <a class="prev" href="{{ $paginator->url($paginator->currentPage()-1) }}">Previous</a>
    @else
        <span class="disabled">Previous</span>
    @endif
    </li>

    <li>
    @if ($paginator->currentPage() !== $paginator->lastPage())
        <a class="next" href="{{ $paginator->url($paginator->currentPage()+1) }}" >Next</a>
    @else
        <span class="disabled">Next</span>
    @endif
    </li>
</ul>
@endif

在我的控制器中，我通过了通常的$model->paginate(10)方法。最后在我看来我想使用Paginator，resources/views/list.blade.php我这样做：

@foreach ($objects as $object)
    // Here we list the object properties
@endforeach
@include("partials.pagination", ['paginator' => $objects])

laravel 5自定义分页器无法正常工作

1 个答案: