在 Django 中获取查询集中的对象计数 [英] Getting a count of objects in a queryset in django

问题描述

如何为数据库中的对象计数添加字段.我有以下型号:

How can I add a field for the count of objects in a database. I have the following models:

class Item(models.Model):
    name = models.CharField()

class Contest(models.Model);
    name = models.CharField()

class Votes(models.Model):
    user = models.ForeignKey(User)
    item = models.ForeignKey(Item)
    contest = models.ForeignKey(Contest)
    comment = models.TextField()

要查找对竞赛 A 的投票，我在视图中使用以下查询

To find the votes for contestA I am using the following query in my view

current_vote = Item.objects.filter(votes__contest=contestA)

这将分别返回一个包含所有投票的查询集，但我想获得每个项目的票数，有人知道我该怎么做吗?谢谢

This returns a queryset with all of the votes individually but I want to get the count votes for each item, anyone know how I can do that? thanks

推荐答案

要获得特定项目的投票数，您可以使用:

To get the number of votes for a specific item, you would use:

vote_count = Item.objects.filter(votes__contest=contestA).count()

如果您想对特定比赛中的投票分布进行细分，我会执行以下操作:

If you wanted a break down of the distribution of votes in a particular contest, I would do something like the following:

contest = Contest.objects.get(pk=contest_id)
votes   = contest.votes_set.select_related()

vote_counts = {}

for vote in votes:
  if not vote_counts.has_key(vote.item.id):
    vote_counts[vote.item.id] = {
      'item': vote.item,
      'count': 0
    }

  vote_counts[vote.item.id]['count'] += 1

这将创建将项目映射到投票数的字典.这不是执行此操作的唯一方法，但它对数据库命中的影响很小，因此运行速度非常快.

This will create dictionary that maps items to number of votes. Not the only way to do this, but it's pretty light on database hits, so will run pretty quickly.

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