或者，如果您能够使用外部库，您可以使用filehound。它支持回调，承诺和同步调用。

使用承诺:

const Filehound = require('filehound');

Filehound.create()
  .path("MyFolder")
  .directory() // only search for directories
  .find()
  .then((subdirectories) => {
    console.log(subdirectories);
  });

使用回调函数:

const Filehound = require('filehound');

Filehound.create()
  .path("MyFolder")
  .directory()
  .find((err, subdirectories) => {
    if (err) return console.error(err);

    console.log(subdirectories);
  });

同步调用:

const Filehound = require('filehound');

const subdirectories = Filehound.create()
  .path("MyFolder")
  .directory()
  .findSync();

console.log(subdirectories);

欲了解更多信息(和示例)，请查看文档:https://github.com/nspragg/filehound

声明:我是作者。

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})); 
            }
        }
    });
}

使用路径列出目录。

function getDirectories(path) {
  return fs.readdirSync(path).filter(function (file) {
    return fs.statSync(path+'/'+file).isDirectory();
  });
}

你可以使用graph-fs

const {Node} = require("graph-fs");
const directory = new Node("/path/to/directory");

const subDirectories = directory.children.filter(child => child.is.directory);

递归解决方案

我来这里是为了寻找一种方法来获取所有子目录，以及它们的所有子目录，等等。在这个公认的答案的基础上，我写道:

const fs = require('fs');
const path = require('path');

function flatten(lists) {
  return lists.reduce((a, b) => a.concat(b), []);
}

function getDirectories(srcpath) {
  return fs.readdirSync(srcpath)
    .map(file => path.join(srcpath, file))
    .filter(path => fs.statSync(path).isDirectory());
}

function getDirectoriesRecursive(srcpath) {
  return [srcpath, ...flatten(getDirectories(srcpath).map(getDirectoriesRecursive))];
}

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])