如果目录不存在,下面的方法是否正确?
它应该对脚本具有完全的权限,并且其他人可以阅读。
var dir = __dirname + '/upload';
if (!path.existsSync(dir)) {
fs.mkdirSync(dir, 0744);
}
如果目录不存在,下面的方法是否正确?
它应该对脚本具有完全的权限,并且其他人可以阅读。
var dir = __dirname + '/upload';
if (!path.existsSync(dir)) {
fs.mkdirSync(dir, 0744);
}
当前回答
不,原因有很多。
The path module does not have an exists/existsSync method. It is in the fs module. (Perhaps you just made a typo in your question?) The documentation explicitly discourage you from using exists. fs.exists() is an anachronism and exists only for historical reasons. There should almost never be a reason to use it in your own code. In particular, checking if a file exists before opening it is an anti-pattern that leaves you vulnerable to race conditions: another process may remove the file between the calls to fs.exists() and fs.open(). Just open the file and handle the error when it's not there. Since we're talking about a directory rather than a file, this advice implies you should just unconditionally call mkdir and ignore EEXIST. In general, you should avoid the *Sync methods. They're blocking, which means absolutely nothing else in your program can happen while you go to the disk. This is a very expensive operation, and the time it takes breaks the core assumption of node's event loop. The *Sync methods are usually fine in single-purpose quick scripts (those that do one thing and then exit), but should almost never be used when you're writing a server: your server will be unable to respond to anyone for the entire duration of the I/O requests. If multiple client requests require I/O operations, your server will very quickly grind to a halt. The only time I'd consider using *Sync methods in a server application is in an operation that happens once (and only once), at startup. For example, require actually uses readFileSync to load modules. Even then, you still have to be careful because lots of synchronous I/O can unnecessarily slow down your server's startup time. Instead, you should use the asynchronous I/O methods.
所以如果我们把这些建议放在一起,我们会得到这样的结果:
function ensureExists(path, mask, cb) {
if (typeof mask == 'function') { // Allow the `mask` parameter to be optional
cb = mask;
mask = 0o744;
}
fs.mkdir(path, mask, function(err) {
if (err) {
if (err.code == 'EEXIST') cb(null); // Ignore the error if the folder already exists
else cb(err); // Something else went wrong
} else cb(null); // Successfully created folder
});
}
我们可以这样使用它:
ensureExists(__dirname + '/upload', 0o744, function(err) {
if (err) // Handle folder creation error
else // We're all good
});
当然,这并不能解释边缘情况,比如
如果在程序运行时删除文件夹会发生什么?(假设在启动过程中只检查它是否存在一次) 如果文件夹已经存在,但是权限错误,会发生什么?
其他回答
使用Node.js 10 + ES6:
import path from 'path';
import fs from 'fs';
(async () => {
const dir = path.join(__dirname, 'upload');
try {
await fs.promises.mkdir(dir);
} catch (error) {
if (error.code === 'EEXIST') {
// Something already exists, but is it a file or directory?
const lstat = await fs.promises.lstat(dir);
if (!lstat.isDirectory()) {
throw error;
}
} else {
throw error;
}
}
})();
适用于v10及以上节点
正如一些回答指出的那样,从节点10开始,您可以对mkdir使用recursive:true
还没有指出的是,当使用recursive:true时,如果目录已经存在,mkdir不会返回错误。
所以你可以这样做:
fsNative.mkdir(dirPath,{recursive:true},(err) => {
if(err) {
//note: this does NOT get triggered if the directory already existed
console.warn(err)
}
else{
//directory now exists
}
})
使用承诺
此外,从节点10开始,您可以通过从fs/promises中要求获得所有fs函数的Promise版本
所以把这两件事放在一起,你就得到了这个简单的解决方案:
import * as fs from 'fs/promises';
await fs.mkdir(dirPath, {recursive:true}).catch((err) => {
//decide what you want to do if this failed
console.error(err);
});
//directory now exists
我的解决方案
CommonJS
var fs = require("fs");
var dir = __dirname + '/upload';
// if (!fs.existsSync(dir)) {
// fs.mkdirSync(dir);
// }
if (!fs.existsSync(dir)) {
fs.mkdirSync(dir, {
mode: 0o744,
});
// mode's default value is 0o744
}
ESM
更新包。json配置
{
//...
"type": "module",
//...
}
import fs from "fs";
import path from "path";
// create one custom `__dirname`, because it not exist in es-module env ⚠️
const __dirname = path.resolve();
const dir = __dirname + '/upload';
if (!fs.existsSync(dir)) {
fs.mkdirSync(dir);
}
// OR
if (!fs.existsSync(dir)) {
fs.mkdirSync(dir, {
mode: 0o744,
});
// mode's default value is 0o744
}
refs
https://nodejs.org/api/fs.html#fsexistssyncpath
https://github.com/nodejs/help/issues/2907#issuecomment-671782092
如果子目录不存在,我必须创建子目录。我用了这个:
const path = require('path');
const fs = require('fs');
function ensureDirectoryExists(p) {
//console.log(ensureDirectoryExists.name, {p});
const d = path.dirname(p);
if (d && d !== p) {
ensureDirectoryExists(d);
}
if (!fs.existsSync(d)) {
fs.mkdirSync(d);
}
}
从文档中可以看出,这是你异步(和递归)做这件事的方式:
const fs = require('fs');
const fsPromises = fs.promises;
fsPromises.access(dir, fs.constants.F_OK)
.catch(async() => {
await fs.mkdir(dir, { recursive: true }, function(err) {
if (err) {
console.log(err)
}
})
});