Question

我可以使用一些我不知道的coffeescript语法来简化这个数组差异代码吗？

first  = [{id:1},{id:2},{id:3},{id:4},{id:5}]
second = [{id:3},{id:4},{id:5},{id:6},{id:7}]

first = first.filter (first_element) ->
  second = second.filter (second_element) ->
    if first_element.id == second_element.id
      first_element.remove = second_element.remove = true
    return !second_element.remove?
  return !first_element.remove?

console.log(first.concat second) # [{id:1},{id:2},{id:6},{id:7}]

Answer 1

这个转换为完全相同的javascript：

first  = [{id:1},{id:2},{id:3},{id:4},{id:5}]
second = [{id:3},{id:4},{id:5},{id:6},{id:7}]

first = first.filter (first_element) ->
  second = second.filter (second_element) ->
    first_element.remove = second_element.remove = yes if first_element.id is second_element.id
    !second_element.remove?
  !first_element.remove?

console.log(first.concat second) # [{id:1},{id:2},{id:6},{id:7}]

删除了明确的return，if放在了行的末尾。 yes替换为true，==替换为is。

虽然行不短，但它保留了数组中的对象：

first = first.filter (firstElement) ->
  keepFirst = yes
  second = second.filter (secondElement) ->
    keepFirst = (keepSecond  =  secondElement.id isnt firstElement.id) and keepFirst
    keepSecond
  keepFirst

请注意，您不能使用and=运算符，因为它短暂循环。

使用语法糖简化数组diff代码

1 个答案: