Question

我正在创建Backbone插件，并假设我有一个CoffeeScript代码，如下所示。

((root, factory) ->
  if typeof exports is "object" and typeof require is "function"
    # CommonJS
    module.exports = factory(require("backbone"))
  else if typeof define is "function" and define.amd
    # AMD
    define [
      "backbone"
    ], (Backbone) ->
      factory Backbone or root.Backbone
  else
    # Browser globals
    factory Backbone
) @, (Backbone) ->
  console.log Backbone # here will be main code.

然后我编译，结果是：

(function() {
  (function(root, factory) {
    if (typeof exports === "object" && typeof require === "function") {
      return module.exports = factory(require("backbone"));
    } else if (typeof define === "function" && define.amd) {
      return define(["backbone"], function(Backbone) {
        return factory(Backbone || root.Backbone);
      }); 
    } else {
      return factory(Backbone);
    }   
  })(this, function(Backbone) {
    return console.log(Backbone);
  }); 

}).call(this);

但是我想潜水为2个CoffeeScript文件（或任意数量的文件），然后用grunt-contrib-coffee连接它们，但编译的JS应该是相同的结果。可能吗？你有什么想法吗？

a.coffee：

((root, factory) ->
  if typeof exports is "object" and typeof require is "function"
    # CommonJS
    module.exports = factory(require("backbone"))
  else if typeof define is "function" and define.amd
    # AMD
    define [
      "backbone"
    ], (Backbone) ->
      factory Backbone or root.Backbone
  else
    # Browser globals
    factory Backbone
) @, (Backbone) ->

b.coffee：

console.log Backbone

更新

根据Kursion的想法，我以某种方式实现了我想做的事，尽管这很棘手; P

grunt indent - ＆gt; grunt concat - ＆gt; grunt coffee

  grunt.initConfig
    indent:
      scripts:
        src: ['src/main.coffee']
        dest: 'tmp/'
        options:
          style: 'space'
          size: 2
          change: 1

    concat:
      sources:
        options:
          separator: ''
        src: [
          'src/entry.coffee'
          'tmp/main.coffee'
        ]   
        dest: 'tmp/example.coffee'

    coffee:
      compile:
        files:
          'lib/example.js': 'tmp/example.coffee'

Answer 1

由于与CoffeeScript相关的缩进问题，这很难。

你可以做的是连接这两个文件的脚本，但你需要尊重第二个文件中的缩进。

连接后，您可以编译CoffeeScript源。

<强> a.coffee

((root, factory) ->
  if typeof exports is "object" and typeof require is "function"
    # CommonJS
    module.exports = factory(require("backbone"))
  else if typeof define is "function" and define.amd
    # AMD
    define [
      "backbone"
    ], (Backbone) ->
      factory Backbone or root.Backbone
  else
    # Browser globals
    factory Backbone
) @, (Backbone) ->

<强> b.coffee

    console.log Backbone # With indentation

两个文件的连接以便：的 glued.coffee

((root, factory) -> if typeof exports is "object" and typeof require is "function" # CommonJS module.exports = factory(require("backbone")) else if typeof define is "function" and define.amd # AMD define [ "backbone" ], (Backbone) -> factory Backbone or root.Backbone else # Browser globals factory Backbone ) @, (Backbone) -> console.log Backbone # With indentation

然后编译：{{1}}

如何在CoffeeScript中编译匿名函数

更新

1 个答案: