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

const fs = require("fs");
fs.rmdir("./test", { recursive: true }, (err) => {
  if (err) {
    console.error(err);
  }
});

提供递归:true选项。并且它将递归地删除给定路径下的所有文件和目录。(假设test是根目录)

其他回答

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。

只需使用rmdir模块!这很简单。

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 });
}

//不使用任何第三方lib

const fs = require('fs');
var FOLDER_PATH = "./dirname";
var files = fs.readdirSync(FOLDER_PATH);
files.forEach(element => {
    fs.unlinkSync(FOLDER_PATH + "/" + element);
});
fs.rmdirSync(FOLDER_PATH);

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/**']);
});