model - 如何在 Rails 中建模 "appointments"？

Question

这是我面临的情况:

可以安排约会:

今天

一周中的某个时间

在特定日期

因此，每个约会“类型”的属性可能不同。

我正在考虑这些模型并将其与 STI 一起使用，但我不确定我是否走在正确的轨道上:

class Appointment < ActiveRecord::Base
class TodayAppointment < Appointment
class WeekAppointment < Appointment
class SpecificDateAppointment < Appointment

table :

string,   Type      #type of the appointment (TodayAppointment, WeekAppointment...)
datetime, When      #data used when type field is "SpecificDateAppointment"
string,   Something #used when type field is "TodayAppointment"

对此进行建模的最佳方法是什么？

这是单表继承的好选择吗？

更新

感谢@Mike，@SpyrosP 到目前为止的帮助。我想出了下面的选项。

这些是数据库表的“ View ”以及它们的外观。

哪一个看起来最合适？

------------------------------------------------------------------------
Option A--(Polymorphic Association)
------------------------------------------------------------------------
|patients               |   day_appointments    |   week_appointments
|   appointment_type    |   data                |   data
|   appointment_id      |                       |
------------------------------------------------------------------------
Option B--(Child references parent) (What is this pattern called?)
------------------------------------------------------------------------
|patients               |   day_appointments    |   week_appointments
|                       |   patient_id          |   patient_id
------------------------------------------------------------------------
Option C--(Polymorphic Association + Single Table Inheritance of appointments)
------------------------------------------------------------------------
|patients               |   appointments        |
|   appointment_type    |   type                |
|   appointment_id      |   day_data            |
|                       |   week_data           |
------------------------------------------------------------------------
Option D--(Child references parent + Single Table Inheritance of appointments)
------------------------------------------------------------------------
|patients               |   appointments        |
|                       |   type                |
|                       |   day_data            |
|                       |   patient_id          |
------------------------------------------------------------------------

Answer 1

你很接近，但看起来你可以从使用 Class Table Inheritance 中受益.我的理由是，每种具体类型都有不同的属性。

这是 Vanilla Ruby 中的一些示例代码。我相信这比我能给出的任何描述都更容易理解。

class Appointment
    def persist
        raise "Must be implemented."
    end
end

class TodayAppointment < Appointment
    def persist
        TodayAppointmentMapper save self
    end
end

class WeekAppointment < Appointment
    def persist
        WeekAppointmentMapper save self
    end
end

class Mapper
    def save aAppointment
        raise "Must be implemented."
    end
end

class TodayAppointmentMapper < Mapper
    def save aAppointment
        # Specfic Today Appointment persistence details.
    end
end

class WeekAppointmentMapper < Mapper
    def save aAppointment
        # Specfic Week Appointment persistence details.
    end
end

请注意每种具体类型能够透明地选择适当的映射器。考虑将其与 Dependency Injection 结合使用以便于测试。