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

看起来ncp和扳手都不再维护了。可能最好的选择是使用fs-extra

扳手的开发人员指导用户使用fs-extra,因为他已经弃用了他的库

copySync和moveSync都将复制和移动文件夹,即使他们有文件或子文件夹,你可以很容易地移动或复制文件使用它

const fse = require('fs-extra');

const srcDir = `path/to/file`;
const destDir = `path/to/destination/directory`;
                                 
// To copy a folder or file, select overwrite accordingly
try {
  fs.copySync(srcDir, destDir, { overwrite: true|false })
  console.log('success!')
} catch (err) {
  console.error(err)
}

OR

// To Move a folder or file, select overwrite accordingly
try {
  fs.moveSync(srcDir, destDir, { overwrite: true|false })
  console.log('success!')
} catch (err) {
  console.error(err)
}

其他回答

这可能是一个可能的解决方案使用异步生成器函数和迭代等待循环。这个解决方案包括过滤掉一些目录的可能性,将它们作为可选的第三个数组参数传递。

import path from 'path';
import { readdir, copy } from 'fs-extra';

async function* getFilesRecursive(srcDir: string, excludedDir?: PathLike[]): AsyncGenerator<string> {
  const directoryEntries: Dirent[] = await readdir(srcDir, { withFileTypes: true });
  if (!directoryEntries.length) yield srcDir; // If the directory is empty, return the directory path.
  for (const entry of directoryEntries) {
    const fileName = entry.name;
      const sourcePath = resolvePath(`${srcDir}/${fileName}`);
      if (entry.isDirectory()) {
        if (!excludedDir?.includes(sourcePath)) {
          yield* getFilesRecursive(sourcePath, excludedDir);
        }
      } else {
        yield sourcePath;
      }
  }
}

然后:

for await (const filePath of getFilesRecursive(path, ['dir1', 'dir2'])) {
   await copy(filePath, filePath.replace(path, path2));
}

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

如果你想递归复制源目录的所有内容,那么你需要将递归选项传递为true,并尝试通过fs-extra记录catch以进行同步

因为fs-extra完全替代了fs,所以你不需要导入基本模块

const fs = require('fs-extra');
let sourceDir = '/tmp/src_dir';
let destDir = '/tmp/dest_dir';
try {
  fs.copySync(sourceDir, destDir, { recursive: true })
  console.log('success!')
} catch (err) {
  console.error(err)
}

这段代码可以很好地工作,递归地将任何文件夹复制到任何位置。但它只适用于Windows。

var child = require("child_process");
function copySync(from, to){
    from = from.replace(/\//gim, "\\");
    to = to.replace(/\//gim, "\\");
    child.exec("xcopy /y /q \"" + from + "\\*\" \"" + to + "\\\"");
}

它非常适合我的基于文本的游戏去创造新玩家。

支持符号链接的:

const path = require("path");
const {
  existsSync,
  mkdirSync,
  readdirSync,
  lstatSync,
  copyFileSync,
  symlinkSync,
  readlinkSync,
} = require("fs");

export function copyFolderSync(src, dest) {
  if (!existsSync(dest)) {
    mkdirSync(dest);
  }

  readdirSync(src).forEach((entry) => {
    const srcPath = path.join(src, entry);
    const destPath = path.join(dest, entry);
    const stat = lstatSync(srcPath);

    if (stat.isFile()) {
      copyFileSync(srcPath, destPath);
    } else if (stat.isDirectory()) {
      copyFolderSync(srcPath, destPath);
    } else if (stat.isSymbolicLink()) {
      symlinkSync(readlinkSync(srcPath), destPath);
    }
  });
}