AASM:保安回调的正确语法

Question

下面是我的示例代码：

class Foo < ActiveRecord::Base
  include AASM
  aasm_column :status
  aasm_initial_state :start_state

  aasm_state :start_state
  aasm_state :state_two
  aasm_state :end_state

  aasm_event :move_to_two, :guard => :guard_callback, :after => :after_callback do
    transitions :from => :start_state, :to => :state_two
  end

  def guard_callback
    puts "executing guard callback..."
    false
  end

  def after_callback
    puts "executing after callback..."
  end

这是我的代码外观的一个玩具表示。我只是从guard回调中返回false，以测试不执行转换或后续转换的行为。下面是我在测试中调用的代码

foo = Foo.new
foo.move_to_two!
puts "foo's current status: #{foo.status}"

下面是输出

executing after callback...
foo's current status: state_two

注意守卫从来不会被叫到..。

我把守卫放错地方了吗？我是不是错了，返回false会停止转换？停止转换是否也会导致忽略after回调？或者，无论发生什么，它都会执行after？

如果最后一件事是真的，我如何将状态传递到回调中？

提前感谢，如果您需要更多信息，请让我知道...

jd

Answer 1

好的，我弄明白了(整个“只要你问，你就会找到答案”的事情)...the :guard就像这样进行转换：

aasm_event :move_to_two, :after => :after_callback do
  transitions :from => :start_state, :to => :state_two, :guard => :guard_callback
end