我希望这是一件简单的事情，但我找不到任何东西在那里这样做。
我只想获得给定文件夹/目录内的所有文件夹/目录。
例如:
<MyFolder>
|- SomeFolder
|- SomeOtherFolder
|- SomeFile.txt
|- SomeOtherFile.txt
|- x-directory
我期望得到一个数组:
["SomeFolder", "SomeOtherFolder", "x-directory"]
或者上面的路径，如果它是这样提供的……
那么，有什么东西已经存在了吗?
当前回答
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)
}
}
其他回答
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])
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
}, [])
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);
