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

这个答案不使用像readdirSync或statSync这样的阻塞函数。它不使用外部依赖关系,也不陷入回调地狱的深渊。

相反,我们使用现代JavaScript的便利,如Promises和async-await语法。异步结果是并行处理的;〇不是按顺序

const { readdir, stat } =
  require ("fs") .promises

const { join } =
  require ("path")

const dirs = async (path = ".") =>
  (await stat (path)) .isDirectory ()
    ? Promise
        .all
          ( (await readdir (path))
              .map (p => dirs (join (path, p)))
          )
        .then
          ( results =>
              [] .concat (path, ...results)
          )
    : []

我将安装一个示例包,然后测试我们的函数-

$ npm install ramda
$ node

让我们看看它是如何工作的

> dirs (".") .then (console.log, console.error)

[ '.'
, 'node_modules'
, 'node_modules/ramda'
, 'node_modules/ramda/dist'
, 'node_modules/ramda/es'
, 'node_modules/ramda/es/internal'
, 'node_modules/ramda/src'
, 'node_modules/ramda/src/internal'
]

使用一个通用模块Parallel,我们可以简化dirs -的定义

const Parallel =
  require ("./Parallel")

const dirs = async (path = ".") =>
  (await stat (path)) .isDirectory ()
    ? Parallel (readdir (path))
        .flatMap (f => dirs (join (path, f)))
        .then (results => [ path, ...results ])
    : []

上面使用的Parallel模块是从一组旨在解决类似问题的函数中提取出来的模式。更多解释请参见相关问答。

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("/")

使用glob包,只需要添加一个斜杠来查找目录:

import {promise as glob} from "glob-promise"

const firstLevelFolders = await glob("MyFolder/*/")
const recursiveFolders = await glob("MyFolder/**/")

对于getDirectories的异步版本,你需要async模块:

var fs = require('fs');
var path = require('path');
var async = require('async'); // https://github.com/caolan/async

// Original function
function getDirsSync(srcpath) {
  return fs.readdirSync(srcpath).filter(function(file) {
    return fs.statSync(path.join(srcpath, file)).isDirectory();
  });
}

function getDirs(srcpath, cb) {
  fs.readdir(srcpath, function (err, files) {
    if(err) { 
      console.error(err);
      return cb([]);
    }
    var iterator = function (file, cb)  {
      fs.stat(path.join(srcpath, file), function (err, stats) {
        if(err) { 
          console.error(err);
          return cb(false);
        }
        cb(stats.isDirectory());
      })
    }
    async.filter(files, iterator, cb);
  });
}

以防其他人从网络搜索到这里,并且已经在他们的依赖列表中有Grunt,这个问题的答案变得微不足道。以下是我的解决方案:

/**
 * Return all the subfolders of this path
 * @param {String} parentFolderPath - valid folder path
 * @param {String} glob ['/*'] - optional glob so you can do recursive if you want
 * @returns {String[]} subfolder paths
 */
getSubfolders = (parentFolderPath, glob = '/*') => {
    return grunt.file.expand({filter: 'isDirectory'}, parentFolderPath + glob);
}