这是此答案的较短的同步版本,它可以列出当前目录中的所有目录(是否隐藏):
const { lstatSync, readdirSync } = require('fs')const { join } = require('path')const isDirectory = source => lstatSync(source).isDirectory()const getDirectories = source => readdirSync(source).map(name => join(source, name)).filter(isDirectory)节点10.10.0+的更新
我们可以使用的新
withFileTypes选项
readdirSync来跳过额外的
lstatSync通话:
const { readdirSync } = require('fs')const getDirectories = source => readdirSync(source, { withFileTypes: true }) .filter(dirent => dirent.isDirectory()) .map(dirent => dirent.name)


