在我的节点应用程序中,我需要删除一个目录,其中有一些文件,但fs。Rmdir只适用于空目录。我该怎么做呢?


当前回答

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");

其他回答

只需使用rmdir模块!这很简单。

我到达这里时,试图克服与大口,我正在写进一步达到。

对于gulp-rimraf,已弃用Gulp-clean Gulp-rimraf已弃用,支持delete-files-folders

当要使用del删除文件和文件夹时,应添加/**进行递归删除。

gulp.task('clean', function () {
    return del(['some/path/to/delete/**']);
});

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。

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+)

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!`);
  });
});