是否有一种更简单的方法来复制文件夹及其所有内容,而无需手动执行一系列的fs。readir, fs。readfile, fs。writefile递归?
我只是想知道我是否错过了一个函数,理想情况下是这样工作的:
fs.copy("/path/to/source/folder", "/path/to/destination/folder");
关于这个历史问题。注意fs。Cp和fs。cpSync可以递归复制文件夹,在Node v16+中可用
是否有一种更简单的方法来复制文件夹及其所有内容,而无需手动执行一系列的fs。readir, fs。readfile, fs。writefile递归?
我只是想知道我是否错过了一个函数,理想情况下是这样工作的:
fs.copy("/path/to/source/folder", "/path/to/destination/folder");
关于这个历史问题。注意fs。Cp和fs。cpSync可以递归复制文件夹,在Node v16+中可用
当前回答
挑选包裹时要小心。有些包(如copy-dir)不支持复制长度超过0X1FFFFFE8个字符(约537 MB)的大文件。
它会抛出一些错误,比如:
buffer.js:630 Uncaught Error:不能创建大于0x1fffffe8字符的字符串
在我的一个项目中,我就经历过类似的事情。最终,我不得不改变我正在使用的包并调整大量代码。我想说,这不是一个很愉快的经历。
如果需要多个源副本和多个目标副本,您可以使用better-copy并编写如下内容:
// Copy from multiple source into a directory
bCopy(['/path/to/your/folder1', '/path/to/some/file.txt'], '/path/to/destination/folder');
甚至:
// Copy from multiple source into multiple destination
bCopy(['/path/to/your/folder1', '/path/to/some/file.txt'], ['/path/to/destination/folder', '/path/to/another/folder']);
其他回答
我尝试了fs-extra和copy-dir来递归地复制文件夹。但我希望它能
正常工作(copy-dir抛出一个不合理的错误) 在过滤器中提供两个参数:filepath和filetype (fs-extra不告诉文件类型) 有从目录到子目录的检查和从目录到文件的检查吗
所以我自己写了:
// Node.js module for Node.js 8.6+
var path = require("path");
var fs = require("fs");
function copyDirSync(src, dest, options) {
var srcPath = path.resolve(src);
var destPath = path.resolve(dest);
if(path.relative(srcPath, destPath).charAt(0) != ".")
throw new Error("dest path must be out of src path");
var settings = Object.assign(Object.create(copyDirSync.options), options);
copyDirSync0(srcPath, destPath, settings);
function copyDirSync0(srcPath, destPath, settings) {
var files = fs.readdirSync(srcPath);
if (!fs.existsSync(destPath)) {
fs.mkdirSync(destPath);
}else if(!fs.lstatSync(destPath).isDirectory()) {
if(settings.overwrite)
throw new Error(`Cannot overwrite non-directory '${destPath}' with directory '${srcPath}'.`);
return;
}
files.forEach(function(filename) {
var childSrcPath = path.join(srcPath, filename);
var childDestPath = path.join(destPath, filename);
var type = fs.lstatSync(childSrcPath).isDirectory() ? "directory" : "file";
if(!settings.filter(childSrcPath, type))
return;
if (type == "directory") {
copyDirSync0(childSrcPath, childDestPath, settings);
} else {
fs.copyFileSync(childSrcPath, childDestPath, settings.overwrite ? 0 : fs.constants.COPYFILE_EXCL);
if(!settings.preserveFileDate)
fs.futimesSync(childDestPath, Date.now(), Date.now());
}
});
}
}
copyDirSync.options = {
overwrite: true,
preserveFileDate: true,
filter: function(filepath, type) {
return true;
}
};
还有一个类似的函数mkdirs,它是mkdirp的替代:
function mkdirsSync(dest) {
var destPath = path.resolve(dest);
mkdirsSync0(destPath);
function mkdirsSync0(destPath) {
var parentPath = path.dirname(destPath);
if(parentPath == destPath)
throw new Error(`cannot mkdir ${destPath}, invalid root`);
if (!fs.existsSync(destPath)) {
mkdirsSync0(parentPath);
fs.mkdirSync(destPath);
}else if(!fs.lstatSync(destPath).isDirectory()) {
throw new Error(`cannot mkdir ${destPath}, a file already exists there`);
}
}
}
对于Linux/Unix操作系统,可以使用shell语法
const shell = require('child_process').execSync;
const src = `/path/src`;
const dist = `/path/dist`;
shell(`mkdir -p ${dist}`);
shell(`cp -r ${src}/* ${dist}`);
就是这样!
是的,ncp很酷…
你可能想要/应该承诺它的功能,让它超级酷。当你这样做的时候,把它添加到一个工具文件中以重复使用它。
下面是一个工作版本,它是异步的,并使用承诺。
文件index.js
const {copyFolder} = require('./tools/');
return copyFolder(
yourSourcePath,
yourDestinationPath
)
.then(() => {
console.log('-> Backup completed.')
}) .catch((err) => {
console.log("-> [ERR] Could not copy the folder: ", err);
})
文件tools.js
const ncp = require("ncp");
/**
* Promise Version of ncp.ncp()
*
* This function promisifies ncp.ncp().
* We take the asynchronous function ncp.ncp() with
* callback semantics and derive from it a new function with
* promise semantics.
*/
ncp.ncpAsync = function (sourcePath, destinationPath) {
return new Promise(function (resolve, reject) {
try {
ncp.ncp(sourcePath, destinationPath, function(err){
if (err) reject(err); else resolve();
});
} catch (err) {
reject(err);
}
});
};
/**
* Utility function to copy folders asynchronously using
* the Promise returned by ncp.ncp().
*/
const copyFolder = (sourcePath, destinationPath) => {
return ncp.ncpAsync(sourcePath, destinationPath, function (err) {
if (err) {
return console.error(err);
}
});
}
module.exports.copyFolder = copyFolder;
对于没有fs的旧节点版本。cp,我在紧要关头使用这个来避免需要第三方库:
const fs = require("fs").promises;
const path = require("path");
const cp = async (src, dest) => {
const lstat = await fs.lstat(src).catch(err => false);
if (!lstat) {
return;
}
else if (await lstat.isFile()) {
await fs.copyFile(src, dest);
}
else if (await lstat.isDirectory()) {
await fs.mkdir(dest).catch(err => {});
for (const f of await fs.readdir(src)) {
await cp(path.join(src, f), path.join(dest, f));
}
}
};
// sample usage
(async () => {
const src = "foo";
const dst = "bar";
for (const f of await fs.readdir(src)) {
await cp(path.join(src, f), path.join(dst, f));
}
})();
相对于现有答案的优势(或区别):
异步 忽略符号链接 如果目录已经存在,则不抛出(如果不需要,则不捕获mkdir抛出) 相当简洁的
我是这样做的:
let fs = require('fs');
let path = require('path');
然后:
let filePath = // Your file path
let fileList = []
var walkSync = function(filePath, filelist)
{
let files = fs.readdirSync(filePath);
filelist = filelist || [];
files.forEach(function(file)
{
if (fs.statSync(path.join(filePath, file)).isDirectory())
{
filelist = walkSync(path.join(filePath, file), filelist);
}
else
{
filelist.push(path.join(filePath, file));
}
});
// Ignore hidden files
filelist = filelist.filter(item => !(/(^|\/)\.[^\/\.]/g).test(item));
return filelist;
};
然后调用该方法:
This.walkSync(filePath, fileList)