Question

我的模型结构的简化示例是

class Corporation(models.Model):
    ...

class Division(models.Model):
    corporation = models.ForeignKey(Corporation)

class Department(models.Model):
    division = models.ForeignKey(Division)
    type = models.IntegerField()

现在我想显示一个表，该表显示公司，其中一列将包含某种类型的部门数，例如type=10。当前，这是通过Corporation模型上的帮助程序来实现的，例如，

class Corporation(models.Model):
    ...
    def get_departments_type_10(self):
        return (
            Department.objects
            .filter(division__corporation=self, type=10)
            .count()
        )

这里的问题是，由于N + 1问题，这绝对会破坏性能。

我尝试使用select_related，prefetch_related，annotate和subquery解决此问题，但是我无法获得所需的结果。

理想情况下，查询集中的每个Corporation都应使用整数type_10_count进行注释，该整数应反映该类型部门的数量。

我确定我可以在.extra()中使用原始sql进行操作，但是文档宣布它将不推荐使用（我在Django 1.11上）

编辑：原始sql解决方案示例

corps = Corporation.objects.raw("""
SELECT
*,
(
    SELECT COUNT(*)
    FROM foo_division div ON div.corporation_id = c.id
    JOIN foo_department dept ON dept.division_id = div.id
    WHERE dept.type = 10
) as type_10_count
FROM foo_corporation c
""")

Answer 1

我认为使用Subquery可以通过此代码获得与您提供的SQL类似的SQL

# Get amount of departments with GROUP BY division__corporation [1]
# .order_by() will remove any ordering so we won't get additional GROUP BY columns [2]
departments = Department.objects.filter(type=10).values(
    'division__corporation'
).annotate(count=Count('id')).order_by()

# Attach departments as Subquery to Corporation by Corporation.id.
# Departments are already grouped by division__corporation
# so .values('count') will always return single row with single column - count [3]
departments_subquery = departments.filter(division__corporation=OuterRef('id'))
corporations = Corporation.objects.annotate(
    departments_of_type_10=Subquery(
        departments_subquery.values('count')
    )
)

生成的SQL是

SELECT "corporation"."id", ... (other fields) ...,
  (
    SELECT COUNT("division"."id") AS "count"
    FROM "department"
    INNER JOIN "division" ON ("department"."division_id" = "division"."id") 
    WHERE (
      "department"."type" = 10 AND
      "division"."corporation_id" = ("corporation"."id")
    ) GROUP BY "division"."corporation_id"
  ) AS "departments_of_type_10"
FROM "corporation"

这里有些担心的是，大表子查询的速度可能很慢。但是，数据库查询优化器足够聪明，可以将子查询提升为OUTER JOIN，至少我听说PostgreSQL做到了。

1. GROUP BY using .values and .annotate

2. order_by() problems

3. Subquery

Answer 2

您应该可以使用Case()表达式来执行此操作，以查询具有您要查找的类型的部门的数量：

from django.db.models import Case, IntegerField, Sum, When, Value

Corporation.objects.annotate(
    type_10_count=Sum(
        Case(
            When(division__department__type=10, then=Value(1)),
            default=Value(0),
            output_field=IntegerField()
        )
    )
)

Django条件子查询聚合

2 个答案: