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

这是我的代码看起来像玩具的代表。我只是从警卫回调中返回false来测试NOT执行转换或after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，警卫永远不会被召唤......

我把警卫放在错误的地方吗？我错误地认为返回虚假会阻止过渡吗？暂停转换是否也会导致后回调被忽略？或者无论如何都会一直执行？

如果最后一件事情是真的，我如何将状态传递给该回调？

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“你一问，你找到了答案”的事情）......：警卫继续自己的过渡：

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

AASM：保护回调的正确语法

1 个答案: