沒有方法錯誤（在異步函數內部訪問方法）

Question

我正在開發Zappa應用程序，並且遇到了我認為JavaScript知識的瓶頸。 問題是我找不到在異步函數內部訪問函數的方法。

當我使用CoffeeScript時，我記得使用that = @或( that = this )類型的作品。 我也嘗試了=>而不是->但是沒有成功。

這是錯誤消息： TypeError: Object #<Object> has no method 'convertUploadedImage'

我該如何克服？

問題在於：

        .on "end", () ->
          file.end()
          console.log "log inside"
          console.log that.constructor.name
          that.convertUploadedImage file

預先感謝。

整個代碼（或至少我認為重要的內容）：

{@app} = require('./node_modules/zappajs') ->
  @post "/collage/upload", (req, res) ->
    # We might be uploading two types of files
    #   1. web form 
    #   2. as a link
    #

    # Check for the upload URL to exist
    url = req.body.upload_url # if req.body.upload_url
    console.log "URL is - #{url}"
    # URL is present - Retrieve the proper file
    if url 
      # setting up for download
      http     = require 'http'
      fileName = url.split('/').pop()
      file     = fs.createWriteStream __dirname + "/public/images/uploads/temp/#{fileName}"

      that = @
      console.log that.constructor.name
      # download the file
      request = http.get "#{url}", ( res ) ->
        res.on "data", (data) ->
          file.write data
        .on "end", () ->
          file.end()
          console.log "log inside"
          console.log that.constructor.name
          that.convertUploadedImage file

    # URL is not present
    else
      file = req.files.files[0]
      @convertUploadedImage file

  convertUploadedImage: ( file, deleteOriginal = false ) ->    
    # Checking file type
    console.log file.type

Answer 1

查看文檔的這一部分（ http://coffeescript.org/#fat-arrow ），粗箭頭應該起作用：

  request = http.get "#{url}", ( res ) =>
    res.on "data", (data) ->
      file.write data
    .on "end", () =>
      file.end()
      console.log "log inside"
      console.log that.constructor.name
      @convertUploadedImage file