django - 将 Django 的 Paginator 类与 MongoDB 游标一起使用

Question

是否有扩展 MongoDB 游标对象以使其与 Djangodjango.core.paginator.Paginator类兼容的已知方法？

或者，也许，扩展 Django 的类？

Answer 1

您的临时解决方案（https://gist.github.com/2351079）看起来不错 - 但不是强制光标使用 获取所有结果list()并使用 分页[bottom:top]，而是尝试在光标上显式使用.skip()和.limit()- 它可能会提高性能。

Answer 2

我遇到了同样的问题并实现了我自己的 Paginator 类。这是代码：

from django.core.paginator import Paginator, Page

class MongoPaginator(Paginator):
    """
    Custom subclass of Django's Paginator to work with Mongo cursors.
    """
    def _get_page(self, *args, **kwargs):
        """
        Returns an instance of a single page. Replaced with our custom
        MongoPage class.
        """
        return MongoPage(*args, **kwargs)

    def page(self, number):
        """
        Returns a Page object for the given 1-based page number.
        Important difference to standard Paginator: Creates a clone of the
        cursor so we can get multiple slices.
        """
        number = self.validate_number(number)
        bottom = (number - 1) * self.per_page
        top = bottom + self.per_page
        if top + self.orphans >= self.count:
            top = self.count
        return self._get_page(
            self.object_list.clone()[bottom:top], number, self
        )

class MongoPage(Page):
    """
    Custom Page class for our MongoPaginator. Just makes sure the cursor is
    directly converted to list so that we can use len(object_list).
    """
    def __init__(self, object_list, number, paginator):
        self.object_list = list(object_list)
        self.number = number
        self.paginator = paginator

主要变化是：

• 每个页面都应该获得光标的克隆，因为切片每个光标只工作一次
• 每个页面还应该直接将其转换为列表，以便 len() 工作

这是在视图中使用它的辅助函数：

def get_paginated_cursor(request, cursor, per_page=25, param='page'):
    """
    Helper to deal with some standard pagination. Pass a request and a
    cursor and it will return a paginated version of it.
    """
    paginator = MongoPaginator(cursor, per_page)

    page = request.GET.get(param, 1)
    try:
        cursor = paginator.page(page)
    except PageNotAnInteger:
        # If page is not an integer, deliver first page.
        cursor = paginator.page(1)
    except EmptyPage:
        # If page is out of range (e.g. 9999), deliver last page of results.
        cursor = paginator.page(paginator.num_pages)

    return cursor

然后你可以这样做：

def some_view_function(request):
    messages_cursor = db.messages.find({})
    messages = get_paginated_cursor(request, messages_cursor)