Question

我只是简单的要求(并非简单执行),并概括了如何实现这一要求,而不必对 d进行多次打击,在盘问时没有<代码>.extra()。

Task:
  name = xxx
  status = models.IntegerField(choices=some_choices)
  project = ForeignKey(Project)

Project:
  name = xxx
  code = xxx

Projects contain Tasks which got various statuses. (Assume status=3 is Completed) Now, I want to list out all projects with their total tasks and completed tasks, like below

Project 1, total_tasks=5, completed_tasks=2
Project 1, total_tasks=2, completed_tasks=1

我能够拿到全部的“信使”,但还没有完成——信使,因为它需要说明条件。是否有办法?

Answer 1

I don t know if it will help, but you can write your own custom annotation objects. I ve just done it though without the conditional part. I based my solution on this link: http://www.voteruniverse.com/Members/jlantz/blog/conditional-aggregates-in-django

但却没有使用这个例子。相反,我审视了集权法,并扩大了苏姆和大标本身。

Answer 2

这一特征是new in Django 1.8.。

参考:。

这有可能:

from django.db.models.aggregates import Count
from django.db.models.expressions import F, Value, Case, When

projects = Project.objects.annotate(
        total_tasks=Count( task__pk ),
        completed_tasks=Count(Case(
           When(status=3, then=F( task__pk )),
           output_field=IntegerField()
        ))).all()

Answer 3

如果你不考虑更多的询问,你可以提出两个问题,而不是一个问题。第一种办法可以让你获得全部票价,第二种可打上<条码>tasks_status,从而使你获得完整的票价。

from django.db.models import Count
all_projects = Project.objects.all()
total_counts = all_projects.annotate(count = Count( tasks ))
completed_counts = all_projects.filter(tasks_status = 3).annotate(count = Count( tasks ))

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