AASM：保护回调的正确语法

Question

这是我的示例代码：

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

这是我的代码的玩具表示。我只是从守卫回调返回 false 来测试不执行转换或之后的行为。这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}"

这是输出

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

请注意，守卫永远不会被调用...

我是否将守卫放在错误的位置？我是否错误地认为返回 false 会停止转换？停止转换是否也会导致 after 回调被忽略？或者无论如何它都会执行 after ？

如果最后一件事是真的，我如何将状态传递到该回调中？

提前致谢，如果您需要更多信息，请告诉我...

jd

Answer 1

好吧，我想通了（整个“只要你问，你就会找到答案”）...... :guard 本身就这样进行转换：

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