使用fs-extra，承诺async fs调用，以及新的await async语法:

const fs = require("fs-extra");

async function getDirectories(path){
    let filesAndDirectories = await fs.readdir(path);

    let directories = [];
    await Promise.all(
        filesAndDirectories.map(name =>{
            return fs.stat(path + name)
            .then(stat =>{
                if(stat.isDirectory()) directories.push(name)
            })
        })
    );
    return directories;
}

let directories = await getDirectories("/")

使用glob包，只需要添加一个斜杠来查找目录:

import {promise as glob} from "glob-promise"

const firstLevelFolders = await glob("MyFolder/*/")
const recursiveFolders = await glob("MyFolder/**/")

这个答案不使用像readdirSync或statSync这样的阻塞函数。它不使用外部依赖关系，也不陷入回调地狱的深渊。

相反，我们使用现代JavaScript的便利，如Promises和async-await语法。异步结果是并行处理的;〇不是按顺序

const { readdir, stat } =
  require ("fs") .promises

const { join } =
  require ("path")

const dirs = async (path = ".") =>
  (await stat (path)) .isDirectory ()
    ? Promise
        .all
          ( (await readdir (path))
              .map (p => dirs (join (path, p)))
          )
        .then
          ( results =>
              [] .concat (path, ...results)
          )
    : []

我将安装一个示例包，然后测试我们的函数-

$ npm install ramda
$ node

让我们看看它是如何工作的

> dirs (".") .then (console.log, console.error)

[ '.'
, 'node_modules'
, 'node_modules/ramda'
, 'node_modules/ramda/dist'
, 'node_modules/ramda/es'
, 'node_modules/ramda/es/internal'
, 'node_modules/ramda/src'
, 'node_modules/ramda/src/internal'
]

使用一个通用模块Parallel，我们可以简化dirs -的定义

const Parallel =
  require ("./Parallel")

const dirs = async (path = ".") =>
  (await stat (path)) .isDirectory ()
    ? Parallel (readdir (path))
        .flatMap (f => dirs (join (path, f)))
        .then (results => [ path, ...results ])
    : []

上面使用的Parallel模块是从一组旨在解决类似问题的函数中提取出来的模式。更多解释请参见相关问答。

这应该做到:

CoffeeScript(同步)

fs = require 'fs'

getDirs = (rootDir) ->
    files = fs.readdirSync(rootDir)
    dirs = []

    for file in files
        if file[0] != '.'
            filePath = "#{rootDir}/#{file}"
            stat = fs.statSync(filePath)

            if stat.isDirectory()
                dirs.push(file)

    return dirs

CoffeeScript(异步)

fs = require 'fs'

getDirs = (rootDir, cb) ->
    fs.readdir rootDir, (err, files) ->
        dirs = []

        for file, index in files
            if file[0] != '.'
                filePath = "#{rootDir}/#{file}"
                fs.stat filePath, (err, stat) ->
                    if stat.isDirectory()
                        dirs.push(file)
                    if files.length == (index + 1)
                        cb(dirs)

JavaScript(异步)

var fs = require('fs');
var getDirs = function(rootDir, cb) { 
    fs.readdir(rootDir, function(err, files) { 
        var dirs = []; 
        for (var index = 0; index < files.length; ++index) { 
            var file = files[index]; 
            if (file[0] !== '.') { 
                var filePath = rootDir + '/' + file; 
                fs.stat(filePath, function(err, stat) {
                    if (stat.isDirectory()) { 
                        dirs.push(this.file); 
                    } 
                    if (files.length === (this.index + 1)) { 
                        return cb(dirs); 
                    } 
                }.bind({index: index, file: file})); 
            }
        }
    });
}

这个答案的CoffeeScript版本，有适当的错误处理:

fs = require "fs"
{join} = require "path"
async = require "async"

get_subdirs = (root, callback)->
    fs.readdir root, (err, files)->
        return callback err if err
        subdirs = []
        async.each files,
            (file, callback)->
                fs.stat join(root, file), (err, stats)->
                    return callback err if err
                    subdirs.push file if stats.isDirectory()
                    callback null
            (err)->
                return callback err if err
                callback null, subdirs

取决于async

或者，使用一个模块! (所有东西都有模块。[引文需要])

另一种递归方法

感谢Mayur了解我的withFileTypes。我写了下面的代码来递归地获取特定文件夹的文件。可以很容易地修改它以只获取目录。

const getFiles = (dir, base = '') => readdirSync(dir, {withFileTypes: true}).reduce((files, file) => {
    const filePath = path.join(dir, file.name)
    const relativePath = path.join(base, file.name)
    if(file.isDirectory()) {
        return files.concat(getFiles(filePath, relativePath))
    } else if(file.isFile()) {
        file.__fullPath = filePath
        file.__relateivePath = relativePath
        return files.concat(file)
    }
}, [])