Question

我将这一小段代码与node.js混合在一起，它允许我创建一个无序的目录列表。我想要的是父循环函数暂停并等待子循环完成然后继续代码。

这是我的代码：

listFolders = (dir, callback) ->
  readDir = "#{__dirname}/public/#{dir}"
  fs.readdir readDir, (err, files) ->
  files.forEach (file) ->
    dirname = file
    path = "#{dir}/#{file}"
    readPath = "#{readDir}/#{file}"

    fs.stat readPath, (err, stat) ->
      if stat && stat.isDirectory()
        console.log "<li class='folder' data-path='#{dir}' data-name='#{dir}/#{dirname}'><span>#{dirname}</span><ul>"
        listFolders path, () ->
          console.log "</ul></li>"
      else if stat && stat.isFile()
        console.log "<li class='file' data-path='#{dir}' data-filename='#{dirname}'><span>#{dirname}</span></li>"

  return callback

正如你所看到的，我试图给函数一个回调但是当我使用它时

listFolders path, () ->
  console.log "</ul></li>"

控制台日志永远不会运行，也不会暂停我的代码。

我正在扫描的文件夹的输出是

<li class='file' data-path='/user_projects/1' data-filename='.DS_Store'><span>.DS_Store</span></li>
<li class='folder' data-path='/user_projects/1' data-name='/user_projects/1/assets'><span>assets</span><ul>
<li class='file' data-path='/user_projects/1' data-filename='index.html'><span>index.html</span></li>
<li class='folder' data-path='/user_projects/1/assets' data-name='/user_projects/1/assets/css'><span>css</span><ul>
<li class='file' data-path='/user_projects/1/assets' data-filename='test.php'><span>test.php</span></li>
<li class='file' data-path='/user_projects/1/assets/css' data-filename='main.css'><span>main.css</span></li>

它应该是

<li class='file' data-path='/user_projects/1' data-filename='.DS_Store'><span>.DS_Store</span></li>
<li class='folder' data-path='/user_projects/1' data-name='/user_projects/1/assets'><span>assets</span><ul>
    <li class='folder' data-path='/user_projects/1/assets' data-name='/user_projects/1/assets/css'><span>css</span><ul>
       <li class='file' data-path='/user_projects/1/assets' data-filename='test.php'><span>test.php</span></li>
       <li class='file' data-path='/user_projects/1/assets/css' data-filename='main.css'><span>main.css</span></li>
    </ul></li>
</ul></li>
<li class='file' data-path='/user_projects/1' data-filename='index.html'><span>index.html</span></li>

我已经忍受了这个问题一段时间了，并希望有人帮助我。非常感谢您的帮助。

＆lt;＆gt;＆gt;＆gt;＆gt;＆gt;＆gt

我现在使用下面提供的代码

listFolders = (dir, callback) ->
  readDir = "#{__dirname}/public/#{dir}"
  fs.readdir readDir, (err, files) ->
    async.forEachSeries files, (file, callback) ->
      dirname = file
      path = "#{dir}/#{file}"
      readPath = "#{readDir}/#{file}"

      fs.stat readPath, (err, stat) ->
        if stat && stat.isDirectory()
          console.log "<li class='folder'><span>#{dirname}</span><ul>"
          listFolders path, () ->
            console.log "</ul></li>"
            callback()
        else if stat && stat.isFile()
          console.log "<li class='file'><span>#{dirname}</span></li>"
          callback()
  , callback

我的输出不是我想要的蚂蚁而是这个（关闭'ul'和'li'没有显示）：

<li class='file' data-path='/user_projects/1' data-filename='.DS_Store'><span>.DS_Store</span></li>
<li class='folder'><span>assets</span><ul>
    <li class='folder'><span>css</span><ul>
        <li class='file'><span>main.css</span></li>

我目录中的文件夹应该显示为

- .ds_store
- asstets
  - css
    - main.css
  - test.php
- index.html

Answer 1

你正在传递一个函数callback，但你永远不会调用它。你刚回来了。您可能希望查看async.js之类的内容。然后你会编写类似于：

的代码

listFolders = (dir, callback) ->
  readDir = "#{__dirname}/public/#{dir}"
  fs.readdir readDir, (err, files) ->
    async.forEachSeries files, (file, callback) ->
      dirname = file
      path = "#{dir}/#{file}"
      readPath = "#{readDir}/#{file}"

      fs.stat readPath, (err, stat) ->
        if stat && stat.isDirectory()
          console.log "<li class='folder' data-path='#{dir}' data-name='#{dir}/#{dirname}'><span>#{dirname}</span><ul>"
          listFolders path, () ->
            console.log "</ul></li>"
            callback()
        else if stat && stat.isFile()
          console.log "<li class='file' data-path='#{dir}' data-filename='#{dirname}'><span>#{dirname}</span></li>"
          callback()
  , callback

如何在CoffeeScript中进行回调

1 个答案: