短小、现代、高效:

import {readdir} from 'node:fs/promises'
import {join} from 'node:path'

const deepReadDir = async (dirPath) => await Promise.all(
  (await readdir(dirPath, {withFileTypes: true})).map(async (dirent) => {
    const path = join(dirPath, dirent.name)
    return dirent.isDirectory() ? await deepReadDir(path) : path
  }),
)

特别感谢函数提示使用{withFileTypes: true}。

这将自动将每个嵌套路径折叠成一个新的嵌套数组。例如，如果:

await deepReadDir('src')

返回如下内容:

[
  [
    'src/client/api.js',
    'src/client/http-constants.js',
    'src/client/index.html',
    'src/client/index.js',
    [ 'src/client/res/favicon.ico' ],
    'src/client/storage.js'
  ],
  [ 'src/crypto/keygen.js' ],
  'src/discover.js',
  [
    'src/mutations/createNewMutation.js',
    'src/mutations/newAccount.js',
    'src/mutations/transferCredit.js',
    'src/mutations/updateApp.js'
  ],
  [
    'src/server/authentication.js',
    'src/server/handlers.js',
    'src/server/quick-response.js',
    'src/server/server.js',
    'src/server/static-resources.js'
  ],
  [ 'src/util/prompt.js', 'src/util/safeWriteFile.js' ],
  'src/util.js'
]

但如果你想，你可以很容易地把它压平:

(await deepReadDir('src')).flat(Number.POSITIVE_INFINITY)

[
  'src/client/api.js',
  'src/client/http-constants.js',
  'src/client/index.html',
  'src/client/index.js',
  'src/client/res/favicon.ico',
  'src/client/storage.js',
  'src/crypto/keygen.js',
  'src/discover.js',
  'src/mutations/createNewMutation.js',
  'src/mutations/newAccount.js',
  'src/mutations/transferCredit.js',
  'src/mutations/updateApp.js',
  'src/server/authentication.js',
  'src/server/handlers.js',
  'src/server/quick-response.js',
  'src/server/server.js',
  'src/server/static-resources.js',
  'src/util/prompt.js',
  'src/util/safeWriteFile.js',
  'src/util.js'
]

我喜欢上面chjj的答案，如果没有那个开始，我就无法创建我的并行循环版本。

var fs = require("fs");

var tree = function(dir, done) {
  var results = {
        "path": dir
        ,"children": []
      };
  fs.readdir(dir, function(err, list) {
    if (err) { return done(err); }
    var pending = list.length;
    if (!pending) { return done(null, results); }
    list.forEach(function(file) {
      fs.stat(dir + '/' + file, function(err, stat) {
        if (stat && stat.isDirectory()) {
          tree(dir + '/' + file, function(err, res) {
            results.children.push(res);
            if (!--pending){ done(null, results); }
          });
        } else {
          results.children.push({"path": dir + "/" + file});
          if (!--pending) { done(null, results); }
        }
      });
    });
  });
};

module.exports = tree;

我也创建了一个Gist。欢迎评论。我仍然在NodeJS领域起步，所以这是我希望学到更多的一种方式。

这是另一个实现。上述解决方案都没有任何限制，因此如果您的目录结构很大，它们都会崩溃并最终耗尽资源。

var async = require('async');
var fs = require('fs');
var resolve = require('path').resolve;

var scan = function(path, concurrency, callback) {
    var list = [];

    var walker = async.queue(function(path, callback) {
        fs.stat(path, function(err, stats) {
            if (err) {
                return callback(err);
            } else {
                if (stats.isDirectory()) {
                    fs.readdir(path, function(err, files) {
                        if (err) {
                            callback(err);
                        } else {
                            for (var i = 0; i < files.length; i++) {
                                walker.push(resolve(path, files[i]));
                            }
                            callback();
                        }
                    });
                } else {
                    list.push(path);
                    callback();
                }
            }
        });
    }, concurrency);

    walker.push(path);

    walker.drain = function() {
        callback(list);
    }
};

使用50的并发工作得非常好，并且几乎和小型目录结构的简单实现一样快。

TypeScript中基于承诺的递归解决方案，使用Array.flat()处理嵌套返回。

import { resolve } from 'path'
import { Dirent } from 'fs'
import * as fs from 'fs'

function getFiles(root: string): Promise<string[]> {
 return fs.promises
   .readdir(root, { withFileTypes: true })
   .then(dirents => {
      const mapToPath = (r: string) => (dirent: Dirent): string => resolve(r, dirent.name)
      const directoryPaths = dirents.filter(a => a.isDirectory()).map(mapToPath(root))
      const filePaths = dirents.filter(a => a.isFile()).map(mapToPath(root))

     return Promise.all<string>([
       ...directoryPaths.map(a => getFiles(a, include)).flat(),
       ...filePaths.map(a => Promise.resolve(a))
     ]).then(a => a.flat())
  })
}

这是一个简单的同步递归解决方案

const fs = require('fs')

const getFiles = path => {
    const files = []
    for (const file of fs.readdirSync(path)) {
        const fullPath = path + '/' + file
        if(fs.lstatSync(fullPath).isDirectory())
            getFiles(fullPath).forEach(x => files.push(file + '/' + x))
        else files.push(file)
    }
    return files
}

用法:

const files = getFiles(process.cwd())

console.log(files)

您可以异步地编写它，但是没有必要。只需确保输入目录存在并且可以访问。

另一个答案，但这次使用的是TypeScript:

/** * Recursively walk a directory asynchronously and obtain all file names (with full path). * * @param dir Folder name you want to recursively process * @param done Callback function, returns all files with full path. * @param filter Optional filter to specify which files to include, * e.g. for json files: (f: string) => /.json$/.test(f) */ const walk = ( dir: string, done: (err: Error | null, results ? : string[]) => void, filter ? : (f: string) => boolean ) => { let results: string[] = []; fs.readdir(dir, (err: Error, list: string[]) => { if (err) { return done(err); } let pending = list.length; if (!pending) { return done(null, results); } list.forEach((file: string) => { file = path.resolve(dir, file); fs.stat(file, (err2, stat) => { if (stat && stat.isDirectory()) { walk(file, (err3, res) => { if (res) { results = results.concat(res); } if (!--pending) { done(null, results); } }, filter); } else { if (typeof filter === 'undefined' || (filter && filter(file))) { results.push(file); } if (!--pending) { done(null, results); } } }); }); }); };