Question

我目前有类似的工作：

period_registration = PeriodRegistration.count(:conditions => ["created_at >= ?", 30.days.ago.to_date], group: "date(created_at)")

但是，我想做这样的事情：

 period_registration_product = PeriodRegistration.count(:conditions => ["created_at >= ?", 30.days.ago.to_date], group: "period_id.product")

当我这样做时，我在period_registration_product中得不到任何回复。在Period_Registration模型中有product_id时，按产品排序的最佳方法是什么？

更新

@period_registration_product = PeriodRegistration.joins(:period).where("date(created_at) >= ?", 30.days.ago.to_date).group("periods.product_id").count

结果：

SQLite3::SQLException: ambiguous column name: created_at: SELECT COUNT(*) AS count_all, periods.product_id AS periods_product_id FROM "period_registrations" INNER JOIN "periods" ON "periods"."id" = "period_registrations"."period_id" WHERE (date(created_at) >= '2012-07-25') GROUP BY periods.product_id

我是否需要指定哪个created_at，我想查看period或period_registration？

Answer 1

尝试此查询

PeriodRegistration.joins(:period).where("period_registrations.created_at >= ?", 30.days.ago.to_date).group("periods.product_id").count

这将返回每个product_id中有多少个对象的哈希

Group By attribute.model可能吗？

1 个答案: