只需使用rmdir模块!这很简单。

return new Promise((resolve, reject) => {
  const fs = require("fs");
  // directory path
  const dir = "your/dir";

  // delete directory recursively <------
  fs.rmdir(dir, { recursive: true }, (err) => {
    if (err) {
      reject(err);
    }
    resolve(`${dir} is deleted!`);
  });
});

const fs = require("fs")
const path = require("path")

let _dirloc = '<path_do_the_directory>'

if (fs.existsSync(_dirloc)) {
  fs.readdir(path, (err, files) => {
    if (!err) {
      for (let file of files) {
        // Delete each file
        fs.unlinkSync(path.join(_dirloc, file))
      }
    }
  })
  // After the 'done' of each file delete,
  // Delete the directory itself.
  if (fs.unlinkSync(_dirloc)) {
    console.log('Directory has been deleted!')
  }
}

如果你更喜欢async/await，你可以使用fs/promises API。

const fs = require('fs/promises');

const removeDir = async (dirPath) => {
  await fs.rm(dirPath, {recursive: true});
}

如果您知道文件夹中单个文件的路径，并希望删除包含该文件的文件夹。

const fs = require('fs/promises');
const path = require('path');

const removeDir = async (filePath) => {
  const { dir } = path.parse(filePath);
  await fs.rm(dir, { recursive: true });
}

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

事实上的包是rimraf，但这里是我的小异步版本:

const fs = require('fs')
const path = require('path')
const Q = require('q')

function rmdir (dir) {
  return Q.nfcall(fs.access, dir, fs.constants.W_OK)
    .then(() => {
      return Q.nfcall(fs.readdir, dir)
        .then(files => files.reduce((pre, f) => pre.then(() => {
          var sub = path.join(dir, f)
          return Q.nfcall(fs.lstat, sub).then(stat => {
            if (stat.isDirectory()) return rmdir(sub)
            return Q.nfcall(fs.unlink, sub)
          })
        }), Q()))
    })
    .then(() => Q.nfcall(fs.rmdir, dir))
}