如何使用Django查询集中的条件注释Count

Question

使用Django ORM,可以做一些像queryset.objects.annotate(Count('queryset_objects', gte=VALUE)).抓住我的漂移？

---

这是一个用于说明可能答案的简单示例:

在Django网站中,内容创建者提交文章,并且普通用户查看(即阅读)所述文章.文章可以发表(即可供所有人阅读),也可以草稿模式.描述这些要求的模型是:

class Article(models.Model):
    author = models.ForeignKey(User)
    published = models.BooleanField(default=False)

class Readership(models.Model):
    reader = models.ForeignKey(User)
    which_article = models.ForeignKey(Article)
    what_time = models.DateTimeField(auto_now_add=True)

我的问题是:如何获得所有发表的文章,按照过去30分钟的独特读者排序？即我想要计算每个发表的文章在过去半小时内获得的不同(独特)视图的数量,然后生成按这些不同视图排序的文章列表.

---

我试过了:

date = datetime.now()-timedelta(minutes=30)
articles = Article.objects.filter(published=True).extra(select = {
  "views" : """
  SELECT COUNT(*)
  FROM myapp_readership
    JOIN myapp_article on myapp_readership.which_article_id = myapp_article.id
  WHERE myapp_readership.reader_id = myapp_user.id
  AND myapp_readership.what_time > %s """ % date,
}).order_by("-views")

这引发了错误:语法错误在"01"或附近(其中"01"是额外的日期时间对象).继续下去并不多.

Answer 1

对于django> = 1.8

使用条件聚合:

from django.db.models import Count, Case, When, IntegerField
Article.objects.annotate(
    numviews=Count(Case(
        When(readership__what_time__lt=treshold, then=1),
        output_field=IntegerField(),
    ))
)

说明: 通过您的文章的正常查询将使用numviews字段进行注释.该字段将被构造为CASE/WHEN表达式,由Count包装,对于读者匹配标准和NULL读者不匹配标准将返回1 .Count将忽略空值并仅计算值.

您将在最近未查看的文章上获得零,并且您可以使用该numviews字段进行排序和过滤.

PostgreSQL背后的查询将是:

SELECT
    "app_article"."id",
    "app_article"."author",
    "app_article"."published",
    COUNT(
        CASE WHEN "app_readership"."what_time" < 2015-11-18 11:04:00.000000+01:00 THEN 1
        ELSE NULL END
    ) as "numviews"
FROM "app_article" LEFT OUTER JOIN "app_readership"
    ON ("app_article"."id" = "app_readership"."which_article_id")
GROUP BY "app_article"."id", "app_article"."author", "app_article"."published"

如果我们只想跟踪唯一的查询,我们可以添加区别Count,并使我们的When子句返回值,我们想要区分.

from django.db.models import Count, Case, When, CharField, F
Article.objects.annotate(
    numviews=Count(Case(
        When(readership__what_time__lt=treshold, then=F('readership__reader')), # it can be also `readership__reader_id`, it doesn't matter
        output_field=CharField(),
    ), distinct=True)
)

这会产生:

SELECT
    "app_article"."id",
    "app_article"."author",
    "app_article"."published",
    COUNT(
        DISTINCT CASE WHEN "app_readership"."what_time" < 2015-11-18 11:04:00.000000+01:00 THEN "app_readership"."reader_id"
        ELSE NULL END
    ) as "numviews"
FROM "app_article" LEFT OUTER JOIN "app_readership"
    ON ("app_article"."id" = "app_readership"."which_article_id")
GROUP BY "app_article"."id", "app_article"."author", "app_article"."published"

对于django <1.8和PostgreSQL

您可以使用它raw来执行由较新版本的django创建的SQL语句.显然没有简单而优化的方法可以在不使用的情况下查询数据raw(即使extra注入了必需的JOIN子句也存在一些问题).

Articles.objects.raw('SELECT'
    '    "app_article"."id",'
    '    "app_article"."author",'
    '    "app_article"."published",'
    '    COUNT('
    '        DISTINCT CASE WHEN "app_readership"."what_time" < 2015-11-18 11:04:00.000000+01:00 THEN "app_readership"."reader_id"'
    '        ELSE NULL END'
    '    ) as "numviews"'
    'FROM "app_article" LEFT OUTER JOIN "app_readership"'
    '    ON ("app_article"."id" = "app_readership"."which_article_id")'
    'GROUP BY "app_article"."id", "app_article"."author", "app_article"."published"')

Answer 2

对于django> = 2.0,您可以在聚合函数中使用带有filter参数的条件聚合:

from datetime import timedelta
from django.utils import timezone
from django.db.models import Count, Q # need import

Article.objects.annotate(
    numviews=Count(
        'readership__reader__id', 
        filter=Q(readership__what_time__gt=timezone.now() - timedelta(minutes=30)), 
        distinct=True
    )
)