Question

 class SocialStudies
      constructor : (@val1,@val2) ->
          console.log 'constructed '+@val1+' | '+@val2
      doAlerts :
          firstAlert : =>
               alert @val1
          secondAlert : =>
               alert @val2

 secondPeriod = new SocialStudies 'git to class!', 'no recess for you!'

 secondPeriod.doAlerts.firstAlert() // error this.val1 is not defined

希望你明白了。我想从方法中的方法集中访问@val1，而胖箭头什么都不做！有谁知道该怎么办？

Answer 1

 class SocialStudies
      constructor : (@val1,@val2) ->
          console.log 'constructed '+@val1+' | '+@val2
          @doAlerts =
            firstAlert : =>
                alert @val1
            secondAlert : =>
                alert @val2

Answer 2

当然，您也可以这样做：

class SocialStudies
  constructor: (@val1, @val2) ->
    @doAlerts = firstAlert: @firstAlert, secondAlert: @secondAlert
  firstAlert: =>
    alert @val1
  secondAlert: =>
    alert @val2

这等同于Keith Nicholas的答案，但允许在继承类的方法中使用 super 关键字，所以你可以这样做：

class AntiSocialStudies extends SocialStudies
  secondAlert: =>
    @val2 += ' no solitary drinking until 3PM.'
    super

secondPeriod = new AntiSocialStudies 'git to class!', 'no recess for you!'
secondPeriod.doAlerts.secondAlert()

Coffeescript类命名空间方法

2 个答案: