邦戈德和“积极记录”这两个接口都相当相似,因此,我认为它们或将会发挥作用。 我需要询问是否符合以下标准。
我需要知道过去10天中某类谷物吃的天数。 谷物每天可以吃10次,但我只需要知道,该日至少吃了1次谷物。 因此,如果今天是第11位,用户在第5次、第7次10次、第8次、第3次、第5次、第5次、第5次、第5次、第5次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次、第3次。
如何做类似于“积极观察”或“刚果”的询问? 基础等级为Cereal,有点名。
我不知道。 Cereal. where (:last_eaten.gt => Annual.today - 10.day) or some such as that in mongoid ? 我在座谈一下,也许不能完全正确。
如果没有适当的模式样本,很难回答。
如果你的模式符合我所铭记的内容(类似于卡坦的答复):
Cereal.where("date > ?", Date.today - 10.days).select("distinct(date)").count
你可能不得不把它 t为bit,但这基本上需要你做些什么。
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