Php Laravel 4到5分页转换

我希望将分页从Laravel 4更改为Laravel 5，我已经解决了使用自定义驱动程序的问题，因此我能够对文件使用Cache:tags

我需要将此代码更改为Laravel 5：

public function buildPaginateQuery(array $params, $page = 1, $perPage = 15)
{
    $query = $this->model->table == 'links' ? $this->model->with('Title') : $this->model;

    $query = $this->appendParams($params, $query);

    $count = $query->cacheTags('count')->remember(2000)->count();

    $totalPages = $count / $perPage;

    $query = $query->skip($perPage * ($page - 1))->take($perPage);

    $query = $query->order(isset($params['order']) && $params['order'] ? $params['order'] : null);

    $query = $query->cacheTags(array($this->model->table, 'pagination'))->remember(2000);

    return array('query' => $query, 'totalPages' => $totalPages, 'totalItems' => $count);
}

我自己也试过，但不知道怎么做，下面的代码是我迄今为止所做的，但当我使用它时，会出现以下错误：

Using $this when not in object context

当前尝试的代码：

/**
 * Builds paginate query with given parameters.
 * 
 * @param  array   $params
 * @param  integer $page
 * @param  integer $perPage
 * 
 * @return array
 */
public function buildPaginateQuery(array $params, $page = 1, $perPage = 15)
{
    $query = $this->model->table == 'links' ? $this->model->with('Title') : $this->model;

    $query = $this->appendParams($params, $query);

    $count = Cache::remember('count', 2000, function() use ($query)
            {
                return $query->count();
            });

    $totalPages = $count / $perPage;

    $query = $query->skip($perPage * ($page - 1))->take($perPage);

    $query = $query->order(isset($params['order']) && $params['order'] ? $params['order'] : null);

    $query = Cache::tags([$this->model->table, 'pagination'])->remember('pagination', 2000, function() use ($query)
            {
                return $query;
            });

    return array('query' => $query, 'totalPages' => $totalPages, 'totalItems' => $count);
}

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错误发生在哪一行？在包含

$query=Cache:：tags（[$this->model->table，'pagination']）->记住（'pagination'，2000，function（）use（$query）{return$query；}）的第几行