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
同步删除文件夹
const fs = require('fs');
const Path = require('path');
const deleteFolderRecursive = function (directoryPath) {
if (fs.existsSync(directoryPath)) {
fs.readdirSync(directoryPath).forEach((file, index) => {
const curPath = path.join(directoryPath, file);
if (fs.lstatSync(curPath).isDirectory()) {
// recurse
deleteFolderRecursive(curPath);
} else {
// delete file
fs.unlinkSync(curPath);
}
});
fs.rmdirSync(directoryPath);
}
};
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。
我希望有一种方法可以做到这一点,而不需要为如此微小和常见的东西添加额外的模块,但这是我能想到的最好的方法。
更新: 现在应该在Windows上工作(测试Windows 10),也应该在Linux/Unix/BSD/Mac系统上工作。
const
execSync = require("child_process").execSync,
fs = require("fs"),
os = require("os");
let removeDirCmd, theDir;
removeDirCmd = os.platform() === 'win32' ? "rmdir /s /q " : "rm -rf ";
theDir = __dirname + "/../web-ui/css/";
// WARNING: Do not specify a single file as the windows rmdir command will error.
if (fs.existsSync(theDir)) {
console.log(' removing the ' + theDir + ' directory.');
execSync(removeDirCmd + '"' + theDir + '"', function (err) {
console.log(err);
});
}
@SharpCoder的答案使用fs.promises的异步版本:
const fs = require('fs');
const afs = fs.promises;
const deleteFolderRecursive = async path => {
if (fs.existsSync(path)) {
for (let entry of await afs.readdir(path)) {
const curPath = path + "/" + entry;
if ((await afs.lstat(curPath)).isDirectory())
await deleteFolderRecursive(curPath);
else await afs.unlink(curPath);
}
await afs.rmdir(path);
}
};
❄️您可以使用graph-fs
directory.delete()
同步删除文件夹
const fs = require('fs');
const Path = require('path');
const deleteFolderRecursive = function (directoryPath) {
if (fs.existsSync(directoryPath)) {
fs.readdirSync(directoryPath).forEach((file, index) => {
const curPath = path.join(directoryPath, file);
if (fs.lstatSync(curPath).isDirectory()) {
// recurse
deleteFolderRecursive(curPath);
} else {
// delete file
fs.unlinkSync(curPath);
}
});
fs.rmdirSync(directoryPath);
}
};
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