我通常不复活旧线程，但这里有很多关于搅动和没有rimraf的答案，这些对我来说似乎都太复杂了。

首先，在现代Node (>= v8.0.0)中，你可以只使用节点核心模块来简化过程，完全异步，并在5行函数中并行化文件的解链接，并且仍然保持可读性:

const fs = require('fs');
const path = require('path');
const { promisify } = require('util');
const readdir = promisify(fs.readdir);
const rmdir = promisify(fs.rmdir);
const unlink = promisify(fs.unlink);

exports.rmdirs = async function rmdirs(dir) {
  let entries = await readdir(dir, { withFileTypes: true });
  await Promise.all(entries.map(entry => {
    let fullPath = path.join(dir, entry.name);
    return entry.isDirectory() ? rmdirs(fullPath) : unlink(fullPath);
  }));
  await rmdir(dir);
};

另一方面，路径遍历攻击的保护不适合此函数，因为

It is out of scope based on the Single Responsibility Principle. Should be handled by the caller not this function. This is akin to the command-line rm -rf in that it takes an argument and will allow the user to rm -rf / if asked to. It would be the responsibility of a script to guard not the rm program itself. This function would be unable to determine such an attack since it does not have a frame of reference. Again that is the responsibility of the caller who would have the context of intent which would provide it a reference to compare the path traversal. Sym-links are not a concern as .isDirectory() is false for sym-links and are unlinked not recursed into.

最后但并非最不重要的是，有一种罕见的竞争条件，即在运行递归时，如果在正确的时间在脚本之外取消链接或删除其中一个条目，则递归可能会出错。由于这种情况在大多数环境中并不典型，因此可能会被忽略。然而，如果需要(对于一些边缘情况)，这个问题可以通过下面这个稍微复杂一点的例子来缓解:

exports.rmdirs = async function rmdirs(dir) {
  let entries = await readdir(dir, { withFileTypes: true });
  let results = await Promise.all(entries.map(entry => {
    let fullPath = path.join(dir, entry.name);
    let task = entry.isDirectory() ? rmdirs(fullPath) : unlink(fullPath);
    return task.catch(error => ({ error }));
  }));
  results.forEach(result => {
    // Ignore missing files/directories; bail on other errors
    if (result && result.error.code !== 'ENOENT') throw result.error;
  });
  await rmdir(dir);
};

编辑:使isDirectory()成为一个函数。最后删除实际目录。修复丢失的递归。

2020的答案

如果你想在npm脚本中完成它，如果你使用npx命令，你不需要预先安装任何第三方包

例如，如果你想在运行npm run clean时删除dist和.cache文件夹，那么只需将此命令添加到package.json中

{
  "scripts": {
    "clean": "npx rimraf dist .cache"
  }
}

它适用于任何操作系统

有一个模块叫做rimraf (https://npmjs.org/package/rimraf)。它提供了与rm -Rf相同的功能

异步用法:

var rimraf = require("rimraf");
rimraf("/some/directory", function () { console.log("done"); });

同步用法:

rimraf.sync("/some/directory");

我写了这个函数叫remove folder。它将递归地删除一个位置中的所有文件和文件夹。它唯一需要的包是异步的。

var async = require('async');

function removeFolder(location, next) {
    fs.readdir(location, function (err, files) {
        async.each(files, function (file, cb) {
            file = location + '/' + file
            fs.stat(file, function (err, stat) {
                if (err) {
                    return cb(err);
                }
                if (stat.isDirectory()) {
                    removeFolder(file, cb);
                } else {
                    fs.unlink(file, function (err) {
                        if (err) {
                            return cb(err);
                        }
                        return cb();
                    })
                }
            })
        }, function (err) {
            if (err) return next(err)
            fs.rmdir(location, function (err) {
                return next(err)
            })
        })
    })
}

❄️您可以使用graph-fs

directory.delete()

另一种替代方法是使用fs-promise模块，该模块提供fs-extra模块的承诺版本

你可以这样写:

const { remove, mkdirp, writeFile, readFile } = require('fs-promise')
const { join, dirname } = require('path')

async function createAndRemove() {
  const content = 'Hello World!'
  const root = join(__dirname, 'foo')
  const file = join(root, 'bar', 'baz', 'hello.txt')

  await mkdirp(dirname(file))
  await writeFile(file, content)
  console.log(await readFile(file, 'utf-8'))
  await remove(join(__dirname, 'foo'))
}

createAndRemove().catch(console.error)

注意:async/await需要最新的nodejs版本(7.6+)