如何在node.js上移动文件(如mv命令shell) ?是否有任何方法,或者我应该读取一个文件,写到一个新文件,并删除旧文件?
当前回答
这个例子摘自:Node.js in Action
move()函数,如果可能,将重命名或返回到复制
var fs = require('fs');
module.exports = function move(oldPath, newPath, callback) {
fs.rename(oldPath, newPath, function (err) {
if (err) {
if (err.code === 'EXDEV') {
copy();
} else {
callback(err);
}
return;
}
callback();
});
function copy() {
var readStream = fs.createReadStream(oldPath);
var writeStream = fs.createWriteStream(newPath);
readStream.on('error', callback);
writeStream.on('error', callback);
readStream.on('close', function () {
fs.unlink(oldPath, callback);
});
readStream.pipe(writeStream);
}
}
其他回答
下面是一个使用util的例子。我如何移动文件a到不同的分区或设备在Node.js?
var fs = require('fs'),
util = require('util');
var is = fs.createReadStream('source_file')
var os = fs.createWriteStream('destination_file');
util.pump(is, os, function() {
fs.unlinkSync('source_file');
});
fs。Rename在同步版本中也可用:
fs.renameSync(oldPath, newPath)
对于8.0.0以上的Node版本使用承诺:
const {promisify} = require('util');
const fs = require('fs');
const {join} = require('path');
const mv = promisify(fs.rename);
const moveThem = async () => {
// Move file ./bar/foo.js to ./baz/qux.js
const original = join(__dirname, 'bar/foo.js');
const target = join(__dirname, 'baz/qux.js');
await mv(original, target);
}
moveThem();
我将把所有涉及的函数(即重命名,复制,断开链接)彼此分离,以获得灵活性,并承诺一切,当然:
const renameFile = (path, newPath) =>
new Promise((res, rej) => {
fs.rename(path, newPath, (err, data) =>
err
? rej(err)
: res(data));
});
const copyFile = (path, newPath, flags) =>
new Promise((res, rej) => {
const readStream = fs.createReadStream(path),
writeStream = fs.createWriteStream(newPath, {flags});
readStream.on("error", rej);
writeStream.on("error", rej);
writeStream.on("finish", res);
readStream.pipe(writeStream);
});
const unlinkFile = path =>
new Promise((res, rej) => {
fs.unlink(path, (err, data) =>
err
? rej(err)
: res(data));
});
const moveFile = (path, newPath, flags) =>
renameFile(path, newPath)
.catch(e => {
if (e.code !== "EXDEV")
throw new e;
else
return copyFile(path, newPath, flags)
.then(() => unlinkFile(path));
});
moveFile只是一个方便的函数,我们可以分别应用这些函数,例如,当我们需要更细粒度的异常处理时。
就像上面的回答中所说的那样,我的2分:copy()方法不应该在没有轻微调整的情况下原样用于复制文件:
function copy(callback) {
var readStream = fs.createReadStream(oldPath);
var writeStream = fs.createWriteStream(newPath);
readStream.on('error', callback);
writeStream.on('error', callback);
// Do not callback() upon "close" event on the readStream
// readStream.on('close', function () {
// Do instead upon "close" on the writeStream
writeStream.on('close', function () {
callback();
});
readStream.pipe(writeStream);
}
复制函数封装在Promise中:
function copy(oldPath, newPath) {
return new Promise((resolve, reject) => {
const readStream = fs.createReadStream(oldPath);
const writeStream = fs.createWriteStream(newPath);
readStream.on('error', err => reject(err));
writeStream.on('error', err => reject(err));
writeStream.on('close', function() {
resolve();
});
readStream.pipe(writeStream);
})
但是,请记住,如果目标文件夹不存在,文件系统可能会崩溃。
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