如何实现不调用count(*)的分页器 [英] How to implement a paginator that doesn't call count(*)
问题描述
我正在django网站上工作,该网站具有MySQL innodb后端.我们的几个表中都有成千上万的记录,这在管理员中引起了一些站点稳定性/性能问题.具体来说,django喜欢在创建分页程序时进行count(*)查询,这会引起很多问题.
I am working on a django website that has a MySQL innodb backend. We have hundreds of thousands of records in several of our tables and this is causing some site stability/performance issues in the admin. Specifically, django likes to make count(*) queries when creating the paginators, and this is causing lots of problems.
使用Django 1.3.x,他们开始允许提供自定义的分页类.因此,我对寻找一种适当地加快或消除这些查询的方法感兴趣.到目前为止,我一直在看这两页: http://code.google.com/p/django-pagination/source/browse/trunk/pagination/paginator.py https://gist.github.com/1094682 并没有真正找到他们正是我要找的东西.任何建议,帮助等.将不胜感激.
With Django 1.3.x, they started to allow for custom pagination classes to be provided. So, I'm interested in finding a way to appropriately speed up or eliminate these queries. So far, I've been looking at these two pages: http://code.google.com/p/django-pagination/source/browse/trunk/pagination/paginator.py https://gist.github.com/1094682 and have not really found them to be what I'm looking for. Any suggestions, help, ect. would be much appreciated.
推荐答案
您可以在分页器中定义_count变量
You can define _count variable in your paginator
paginator = Paginator(QuerySet, 300)
paginator._count = 9000 # or use some query here
这是Django分页器代码的一部分,可帮助您了解此变量的作用以及页数的工作原理
And here is the part of django paginator code to help you understand what this variable do and how page count works
def _get_count(self):
"Returns the total number of objects, across all pages."
if self._count is None:
try:
self._count = self.object_list.count()
except (AttributeError, TypeError):
# AttributeError if object_list has no count() method.
# TypeError if object_list.count() requires arguments
# (i.e. is of type list).
self._count = len(self.object_list)
return self._count
count = property(_get_count)
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