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())
  })
}

There are basically two ways of accomplishing this. In an async environment you'll notice that there are two kinds of loops: serial and parallel. A serial loop waits for one iteration to complete before it moves onto the next iteration - this guarantees that every iteration of the loop completes in order. In a parallel loop, all the iterations are started at the same time, and one may complete before another, however, it is much faster than a serial loop. So in this case, it's probably better to use a parallel loop because it doesn't matter what order the walk completes in, just as long as it completes and returns the results (unless you want them in order).

一个平行循环看起来是这样的:

var fs = require('fs');
var path = require('path');
var walk = function(dir, done) {
  var results = [];
  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) {
      file = path.resolve(dir, file);
      fs.stat(file, function(err, stat) {
        if (stat && stat.isDirectory()) {
          walk(file, function(err, res) {
            results = results.concat(res);
            if (!--pending) done(null, results);
          });
        } else {
          results.push(file);
          if (!--pending) done(null, results);
        }
      });
    });
  });
};

一个串行循环看起来像这样:

var fs = require('fs');
var path = require('path');
var walk = function(dir, done) {
  var results = [];
  fs.readdir(dir, function(err, list) {
    if (err) return done(err);
    var i = 0;
    (function next() {
      var file = list[i++];
      if (!file) return done(null, results);
      file = path.resolve(dir, file);
      fs.stat(file, function(err, stat) {
        if (stat && stat.isDirectory()) {
          walk(file, function(err, res) {
            results = results.concat(res);
            next();
          });
        } else {
          results.push(file);
          next();
        }
      });
    })();
  });
};

并且在你的主目录中测试它(警告:如果你的主目录中有很多东西，结果列表将会非常大):

walk(process.env.HOME, function(err, results) {
  if (err) throw err;
  console.log(results);
});

编辑:改进的示例。

a .看一下文件模块。它有一个叫walk的函数:

文件。步行(开始,回调) 导航文件树，为每个目录调用回调，传入 (null, dirPath, dirs, files)。

这可能是为你准备的!是的，它是异步的。但是，如果需要的话，我认为您必须自己聚合完整的路径。

B.另一种选择，甚至是我的最爱之一:使用unix find来查找。为什么要再做一件已经编程好的事情呢?也许不是你真正需要的，但仍然值得一试:

var execFile = require('child_process').execFile;
execFile('find', [ 'somepath/' ], function(err, stdout, stderr) {
  var file_list = stdout.split('\n');
  /* now you've got a list with full path file names */
});

Find有一个很好的内置缓存机制，使得后续搜索非常快，只要只有少数文件夹被更改。

下面是完整的工作代码。按您的要求。您可以递归地获取所有文件和文件夹。

var recur = function(dir) {
            fs.readdir(dir,function(err,list){
                list.forEach(function(file){
                    var file2 = path.resolve(dir, file);
                    fs.stat(file2,function(err,stats){
                        if(stats.isDirectory()) {
                            recur(file2);
                        }
                        else {
                            console.log(file2);
                        }
                    })
                })
            });
        };
        recur(path);

在路径中给出你想要搜索的目录路径，如"c:\test"

我必须将基于promise的砂光器库添加到列表中。

 var sander = require('sander');
 sander.lsr(directory).then( filenames => { console.log(filenames) } );

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。