Rails模型继承和路由

class User < ActiveRecord::Base 
    has_many :queued_workouts, 
      :conditions => "queue_position IS NOT NULL", 
      :order => "queue_position ASC" 

    has_many :workouts 
end 

class Workout < ActiveRecord::Base 
end 

class QueuedWorkout < Workout 
    # Have to give this its own class because acts_as_list always updates the 
    # queue_position when operating on user.workouts 
    acts_as_list :column => :queue_position, :scope => :user 
end

我有锻炼的路线，但不需要它们用于QueuedWorkouts。每过一段时间，我都遇到了将QueuedWorkout而不是Workout传递到url_for的情况。在我的具体情况下，这发生在WorkoutObserver中。Rails模型继承和路由

现在我做

class WorkoutObserver < ActiveRecord::Observer 
    def after_update(workout) 
    workout = Workout.find(workout.id) if workout.is_a? QueuedWorkout 
    twitter_status = "some stuff " + short_url_for(workout) # Helper method for generating urls outside controllers & views 
    .... 
    end 
end

，当然这是一个可怕的解决方案。有没有更好的方法来告诉Rails路由器为QueuedWorkouts生成锻炼URL？

来源

2011-09-26 Dean Brundage

你可以这样做：

resources :queued_workout, :controller => "workout"

来源

2011-09-27 00:28:03 drummondj

因为我已经切换到这一点，但宁愿有路由器自动复制路线。或跳过路由并在url_for中执行。 –

Rails模型继承和路由

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