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
这在Node.js 10中非常简单:
const Path = require('path');
const FSP = require('fs').promises;
async function copyDir(src,dest) {
const entries = await FSP.readdir(src, {withFileTypes: true});
await FSP.mkdir(dest);
for(let entry of entries) {
const srcPath = Path.join(src, entry.name);
const destPath = Path.join(dest, entry.name);
if(entry.isDirectory()) {
await copyDir(srcPath, destPath);
} else {
await FSP.copyFile(srcPath, destPath);
}
}
}
这里假设dest不存在。
其他回答
内联版本
node -e "const fs=require('fs');const p=require('path');function copy(src, dest) {if (!fs.existsSync(src)) {return;} if (fs.statSync(src).isFile()) {fs.copyFileSync(src, dest);}else{fs.mkdirSync(dest, {recursive: true});fs.readdirSync(src).forEach(f=>copy(p.join(src, f), p.join(dest, f)));}}const args=Array.from(process.argv); copy(args[args.length-2], args[args.length-1]);" dist temp\dest
或者节点16.x+
node -e "const fs=require('fs');const args=Array.from(process.argv); fs.cpSync(args[args.length-2], args[args.length-1], {recursive: true});"
在“节点14.20.0”上测试，但假设它在节点10.x上工作?
来自user8894303和pen的回答:https://stackoverflow.com/a/52338335/458321
如果在包中使用，请务必转义引号。json脚本
package.json:
"scripts": {
"rmrf": "node -e \"const fs=require('fs/promises');const args=Array.from(process.argv); Promise.allSettled(args.map(a => fs.rm(a, { recursive: true, force: true })));\"",
"cp": "node -e \"const fs=require('fs');const args=Array.from(process.argv);if (args.length>2){ fs.cpSync(args[args.length-2], args[args.length-1], {recursive: true});}else{console.log('args missing', args);}\""
"copy": "node -e \"const fs=require('fs');const p=require('path');function copy(src, dest) {if (!fs.existsSync(src)) {return;} if (fs.statSync(src).isFile()) {fs.copyFileSync(src, dest);}else{fs.mkdirSync(dest, {recursive: true});fs.readdirSync(src).forEach(f=>copy(p.join(src, f), p.join(dest, f)));}}const args=Array.from(process.argv);if (args.length>2){copy(args[args.length-2], args[args.length-1]);}else{console.log('args missing', args);}\"",
"mkdir": "node -e \"const fs=require('fs');const args=Array.from(process.argv);fs.mkdirSync(args[args.length-1],{recursive:true});\"",
"clean": "npm run rmrf -- temp && npm run mkdir -- temp && npm run copy -- dist temp"
}
注:RMRF脚本需要14.20节点。X还是12.20.x?
奖金:
deno eval "import { existsSync, mkdirSync, copyFileSync, readdirSync, statSync } from 'node:fs';import { join } from 'node:path';function copy(src, dest) {if (!existsSync(src)) {return;} if (statSync(src).isFile()) {copyFileSync(src, dest);}else{mkdirSync(dest, {recursive: true});readdirSync(src).forEach(f=>copy(join(src, f), join(dest, f)));}}const args=Array.from(Deno.args);copy(args[0], args[1]);" dist temp\dest -- --allow-read --allow-write
Deno支持-> NPM I Deno -bin支持节点中的Deno -bin
有一些模块支持复制文件夹及其内容。最流行的是wrench.js:
// Deep-copy an existing directory
wrench.copyDirSyncRecursive('directory_to_copy', 'location_where_copy_should_end_up');
一个替代方案是node-fs-extra:
fs.copy('/tmp/mydir', '/tmp/mynewdir', function (err) {
if (err) {
console.error(err);
} else {
console.log("success!");
}
}); // Copies directory, even if it has subdirectories or files
解决这个问题最简单的方法是只使用'fs'和'Path'模块和一些逻辑…
如果你只是想设置版本号，根文件夹中的所有文件都复制新名称，即" var v = '您的目录名'"
在文件名的前缀中添加带有文件名的内容。
var fs = require('fs-extra');
var path = require('path');
var c = 0;
var i = 0;
var v = "1.0.2";
var copyCounter = 0;
var directoryCounter = 0;
var directoryMakerCounter = 0;
var recursionCounter = -1;
var Flag = false;
var directoryPath = [];
var directoryName = [];
var directoryFileName = [];
var fileName;
var directoryNameStorer;
var dc = 0;
var route;
if (!fs.existsSync(v)) {
fs.mkdirSync(v);
}
var basePath = path.join(__dirname, v);
function walk(dir) {
fs.readdir(dir, function(err, items) {
items.forEach(function(file) {
file = path.resolve(dir, file);
fs.stat(file, function(err, stat) {
if(stat && stat.isDirectory()) {
directoryNameStorer = path.basename(file);
route = file;
route = route.replace("gd", v);
directoryFileName[directoryCounter] = route;
directoryPath[directoryCounter] = file;
directoryName[directoryCounter] = directoryNameStorer;
directoryCounter++;
dc++;
if (!fs.existsSync(basePath + "/" + directoryName[directoryMakerCounter])) {
fs.mkdirSync(directoryFileName[directoryMakerCounter]);
directoryMakerCounter++;
}
}
else {
fileName = path.basename(file);
if(recursionCounter >= 0) {
fs.copyFileSync(file, directoryFileName[recursionCounter] + "/" + v + "_" + fileName, err => {
if(err) return console.error(err);
});
copyCounter++;
}
else {
fs.copyFileSync(file, v + "/" + v + "_" + fileName, err => {
if(err) return console.error(err);
});
copyCounter++;
}
}
if(copyCounter + dc == items.length && directoryCounter > 0 && recursionCounter < directoryMakerCounter-1) {
console.log("COPY COUNTER: " + copyCounter);
console.log("DC COUNTER: " + dc);
recursionCounter++;
dc = 0;
copyCounter = 0;
console.log("ITEM DOT LENGTH: " + items.length);
console.log("RECURSION COUNTER: " + recursionCounter);
console.log("DIRECOTRY MAKER COUNTER: " + directoryMakerCounter);
console.log(": START RECURSION: " + directoryPath[recursionCounter]);
walk(directoryPath[recursionCounter]); //recursive call to copy sub-folder
}
})
})
});
}
walk('./gd', function(err, data) { // Just pass the root directory which you want to copy
if(err)
throw err;
console.log("done");
})
如果你在Linux上，性能不是问题，你可以使用child_process模块中的exec函数来执行一个Bash命令:
const { exec } = require('child_process');
exec('cp -r source dest', (error, stdout, stderr) => {...});
在某些情况下，我发现这个解决方案比下载整个模块甚至使用fs模块更简洁。
这是我个人的做法:
function copyFolderSync(from, to) {
fs.mkdirSync(to);
fs.readdirSync(from).forEach(element => {
if (fs.lstatSync(path.join(from, element)).isFile()) {
fs.copyFileSync(path.join(from, element), path.join(to, element));
} else {
copyFolderSync(path.join(from, element), path.join(to, element));
}
});
}
它适用于文件夹和文件。
