我已经读了很多https://docs.djangoproject.com/en/1.11/topics/db/aggregation/,但是我仍然遗漏了一些东西。
使用Django 1.11,假设我有以下模型:
class School(models.Model):
pass
class Classroom(models.Model):
school = models.ForeignKey(School, on_delete=models.PROTECT)
active = models.BooleanField()
busy = models.BooleanField()
class Chalkboard(models.Model):
classroom = models.ForeignKey(Classroom, on_delete=models.PROTECT)
class Whiteboard(models.Model):
classroom = models.ForeignKey(Classroom, on_delete=models.PROTECT)我创建了一所学校,有一个教室,有两个白板和两个黑板:
s = School()
s.save()
c = Classroom(school=s, active=True, busy=False)
c.save()
Chalkboard(classroom=c).save()
Chalkboard(classroom=c).save()
Whiteboard(classroom=c).save()
Whiteboard(classroom=c).save()
Whiteboard(classroom=c).save()我想要一个总结,有多少黑板在每所学校是积极的,但不忙。
q = School.objects.filter(
Q(classroom__active=True) & Q(classroom__busy=False)
).annotate(
chalkboard_count=Count('classroom__chalkboard'),
)
q[0].chalkboard_count
2 # as expected现在我想知道黑板和白板。
q = School.objects.filter(
Q(classroom__active=True) & Q(classroom__busy=False)
).annotate(
chalkboard_count=Count('classroom__chalkboard'),
whiteboard_count=Count('classroom__whiteboard'), # added this line
)
q[0].chalkboard_count
6 # expected 2
q[0].whiteboard_count
6 # expected 3如果我将调用链接到注释,就会得到相同的结果。
q = School.objects.filter(
Q(classroom__active=True) & Q(classroom__busy=False)
).annotate(
chalkboard_count=Count('classroom__chalkboard')
).annotate(
whiteboard_count=Count('classroom__whiteboard')
)
q[0].chalkboard_count
6 # expected 2
q[0].whiteboard_count
6 # expected 3一直以来,这些数字都是我所期望的
Chalkboard.objects.count()
2
Whiteboard.objects.count()
3我在这里做错什么了?
发布于 2018-01-10 00:08:20
从你发布的链接中:
组合多个聚合 将多个聚合与注解()组合在一起会产生错误的结果,因为使用的是联接而不是子查询:对于大多数聚合来说,没有办法避免这个问题,然而,Count聚合有一个不同的参数,可能有助于:
Book.objects.annotate(
Count('authors', distinct=True),
Count('store', distinct=True)
)https://stackoverflow.com/questions/48177300
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