使用node.js版本>= v10.13.0, fs. js。readdirSync将返回一个fs数组。如果withFileTypes选项设置为true，则直接对象。

所以你可以用，

const fs = require('fs')

const directories = source => fs.readdirSync(source, {
   withFileTypes: true
}).reduce((a, c) => {
   c.isDirectory() && a.push(c.name)
   return a
}, [])

使用路径列出目录。

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

承诺

import { readdir } from 'fs/promises'

const getDirectories = async source =>
  (await readdir(source, { withFileTypes: true }))
    .filter(dirent => dirent.isDirectory())
    .map(dirent => dirent.name)

回调

import { readdir } from 'fs'

const getDirectories = (source, callback) =>
  readdir(source, { withFileTypes: true }, (err, files) => {
    if (err) {
      callback(err)
    } else {
      callback(
        files
          .filter(dirent => dirent.isDirectory())
          .map(dirent => dirent.name)
      )
    }
  })

Syncronous

import { readdirSync } from 'fs'

const getDirectories = source =>
  readdirSync(source, { withFileTypes: true })
    .filter(dirent => dirent.isDirectory())
    .map(dirent => dirent.name)

函数式编程

const fs = require('fs')
const path = require('path')
const R = require('ramda')

const getDirectories = pathName => {
    const isDirectory = pathName => fs.lstatSync(pathName).isDirectory()
    const mapDirectories = pathName => R.map(name => path.join(pathName, name), fs.readdirSync(pathName))
    const filterDirectories = listPaths => R.filter(isDirectory, listPaths)

    return {
        paths:R.pipe(mapDirectories)(pathName),
        pathsFiltered: R.pipe(mapDirectories, filterDirectories)(pathName)
    }
}

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)
    }
}, [])