因为每个人都应该写自己的，所以我写了一个。

步行(dir, cb, endCb) cb(文件) 零endCb (err |)

脏

module.exports = walk;

function walk(dir, cb, endCb) {
  var fs = require('fs');
  var path = require('path');

  fs.readdir(dir, function(err, files) {
    if (err) {
      return endCb(err);
    }

    var pending = files.length;
    if (pending === 0) {
      endCb(null);
    }
    files.forEach(function(file) {
      fs.stat(path.join(dir, file), function(err, stats) {
        if (err) {
          return endCb(err)
        }

        if (stats.isDirectory()) {
          walk(path.join(dir, file), cb, function() {
            pending--;
            if (pending === 0) {
              endCb(null);
            }
          });
        } else {
          cb(path.join(dir, file));
          pending--;
          if (pending === 0) {
            endCb(null);
          }
        }
      })
    });

  });
}

谁想要一个公认答案的同步替代方案(我知道我做过):

var fs = require('fs');
var path = require('path');
var walk = function(dir) {
    let results = [], err = null, list;
    try {
        list = fs.readdirSync(dir)
    } catch(e) {
        err = e.toString();
    }
    if (err) return err;
    var i = 0;
    return (function next() {
        var file = list[i++];

        if(!file) return results;
        file = path.resolve(dir, file);
        let stat = fs.statSync(file);
        if (stat && stat.isDirectory()) {
          let res = walk(file);
          results = results.concat(res);
          return next();
        } else {
          results.push(file);
           return next();
        }

    })();

};

console.log(
    walk("./")
)

递归-readdir模块具有此功能。

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); } } }); }); }); };

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。

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())
  })
}