需要一种解决方法来过滤 Django 中的相关模型和聚合字段

Question

我为这个问题打开了ticket。

简而言之，这是我的模型：

class Plan(models.Model):
 cap = models.IntegerField()

class Phone(models.Model):
 plan = models.ForeignKey(Plan, related_name='phones')

class Call(models.Model):
 phone = models.ForeignKey(Phone, related_name='calls')
 cost = models.IntegerField()

我想运行这样的查询：

Phone.objects.annotate(total_cost=Sum('calls__cost')).filter(total_cost__gte=0.5*F('plan__cap'))

不幸的是 Django 生成了错误的 SQL：

SELECT "app_phone"."id", "app_phone"."plan_id",
SUM("app_call"."cost") AS "total_cost"
FROM "app_phone"
INNER JOIN "app_plan" ON ("app_phone"."plan_id" = "app_plan"."id")
LEFT OUTER JOIN "app_call" ON ("app_phone"."id" = "app_call"."phone_id")
GROUP BY "app_phone"."id", "app_phone"."plan_id"
HAVING SUM("app_call"."cost") >=  0.5 * "app_plan"."cap"

和错误：

ProgrammingError: column "app_plan.cap" must appear in the GROUP BY clause or be used in an aggregate function
LINE 1: ...."plan_id" HAVING SUM("app_call"."cost") >=  0.5 * "app_plan"....

除了运行原始 SQL 之外，还有其他解决方法吗？

Answer 1

在聚合时，SQL 要求字段中的任何值在组内都是唯一的，或者将该字段包装在聚合函数中，以确保每个组只输出一个值。这里的问题是“app_plan.cap”对于“app_phone.id”和“app_phone.plan_id”的每个组合可能有许多不同的值，所以你需要告诉数据库如何处理这些。

因此，您的结果的有效 SQL 是两种不同的可能性之一，具体取决于您想要的结果。首先，您可以在 GROUP BY 函数中包含 app_plan.cap，这样 (app_phone.id, app_phone.plan_id, app_plan.cap) 的任何不同组合都将是不同的组：

SELECT "app_phone"."id", "app_phone"."plan_id", "app_plan"."cap",
SUM("app_call"."cost") AS "total_cost"
FROM "app_phone"
INNER JOIN "app_plan" ON ("app_phone"."plan_id" = "app_plan"."id")
LEFT OUTER JOIN "app_call" ON ("app_phone"."id" = "app_call"."phone_id")
GROUP BY "app_phone"."id", "app_phone"."plan_id", "app_plan"."cap"
HAVING SUM("app_call"."cost") >=  0.5 * "app_plan"."cap"

诀窍是在“GROUP BY”调用中获取额外的值。我们可以通过滥用“extra”来解决这个问题，尽管这会将表名硬编码为“app_plan”，这是不理想的——如果你愿意，你可以使用 Plan 类以编程方式进行：

Phone.objects.extra({
    "plan_cap": "app_plan.cap"
}).annotate(
    total_cost=Sum('calls__cost')
).filter(total_cost__gte=0.5*F('plan__cap'))

或者，您可以将app_plan.cap 包装在聚合函数中，将其转换为唯一值。聚合函数因数据库提供商而异，但可能包括 AVG、MAX、MIN 等。

SELECT "app_phone"."id", "app_phone"."plan_id",
SUM("app_call"."cost") AS "total_cost",
AVG("app_plan"."cap") AS "avg_cap",
FROM "app_phone"
INNER JOIN "app_plan" ON ("app_phone"."plan_id" = "app_plan"."id")
LEFT OUTER JOIN "app_call" ON ("app_phone"."id" = "app_call"."phone_id")
GROUP BY "app_phone"."id", "app_phone"."plan_id"
HAVING SUM("app_call"."cost") >=  0.5 * AVG("app_plan"."cap")

您可以使用以下方法在 Django 中获得此结果：

Phone.objects.annotate(
    total_cost=Sum('calls__cost'), 
    avg_cap=Avg('plan__cap')
).filter(total_cost__gte=0.5 * F("avg_cap"))

您可能需要考虑更新您留下的错误报告，更清楚地说明您期望的结果——例如，您所追求的有效 SQL。