laravel - 如何使用Laravel 5在Elasticsearch中分页

Question

我是Elasticsearch的新手，我想在其中使用分页。

我正在使用from和size。我现在总共有10条记录，而我正在从0扩展到5。我得到5个结果。但是，如何给链接提供查看页面以及如何增加第二页的记录呢？

我的查询:

       $params = [
                'index' => 'my_index',
                'type' => 'product',
                'body' =>  [
                    "sort" =>[
                                ["default_product_low_price.sale_price" => ["order" => $sort]]
                            ],
                    "from" => 0, "size" =>5,
                    'query'=> $query,
                  ]
                ];
          $response = \Es::Search($params);

这是我的查询现在在哪里可以给出分页链接？

Answer 1

在仓库中

        $params['size'] = $per_page;
$params['from'] = $from;
$params['index'] = config('elastic.Admin_Logs');
$params['type'] = config('elastic.Admin_Type');
$params['body']['sort']['default_product_low_price.sale_price']['order'] = "desc";
$params['body']['query']['filtered']['filter']['bool']['must'][]['match_all'] = [];

$response = \Es::Search($params);
 $access = $response['hits'];
        return $access;

在 Controller 中，我设置了$ per_page和$ from

$per_page = $request->get('limit', 10);
    $from = ($request->get('page', 1) - 1) * $per_page;
    $access = $this->repository->Index($per_page, $from);

    $admin_exceptions = new LengthAwarePaginator(
            $access['hits'],
            $access['total'],
            $per_page,
            Paginator::resolveCurrentPage(),
            ['path' => Paginator::resolveCurrentPath()]);
 return view('adminexception.index', compact('admin_exceptions'))->withInput($request->all());

现在在 View {{!! $ admin_exceptions-> render()!!}}中使用render