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

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("/")

其他回答

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])

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)
    }
}

完全异步的版本与ES6,只有本机包,fs。Promises和async/await,并行执行文件操作:

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

async function listDirectories(rootPath) {
    const fileNames = await fs.promises.readdir(rootPath);
    const filePaths = fileNames.map(fileName => path.join(rootPath, fileName));
    const filePathsAndIsDirectoryFlagsPromises = filePaths.map(async filePath => ({path: filePath, isDirectory: (await fs.promises.stat(filePath)).isDirectory()}))
    const filePathsAndIsDirectoryFlags = await Promise.all(filePathsAndIsDirectoryFlagsPromises);
    return filePathsAndIsDirectoryFlags.filter(filePathAndIsDirectoryFlag => filePathAndIsDirectoryFlag.isDirectory)
        .map(filePathAndIsDirectoryFlag => filePathAndIsDirectoryFlag.path);
}

经过测试,它工作得很好。

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))];
}