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

我写了这个函数用于在目录之间递归地复制(copyFileSync)或移动(renameSync)文件:

// Copy files
copyDirectoryRecursiveSync(sourceDir, targetDir);
// Move files
copyDirectoryRecursiveSync(sourceDir, targetDir, true);


function copyDirectoryRecursiveSync(source, target, move) {
    if (!fs.lstatSync(source).isDirectory())
        return;

    var operation = move ? fs.renameSync : fs.copyFileSync;
    fs.readdirSync(source).forEach(function (itemName) {
        var sourcePath = path.join(source, itemName);
        var targetPath = path.join(target, itemName);

        if (fs.lstatSync(sourcePath).isDirectory()) {
            fs.mkdirSync(targetPath);
            copyDirectoryRecursiveSync(sourcePath, targetPath);
        }
        else {
            operation(sourcePath, targetPath);
        }
    });
}

其他回答

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

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。

目前最上面的答案可以大大简化。

const path = require('path');
const fs = require('fs');

function recursiveCopySync(source, target) {
  if (fs.lstatSync(source).isDirectory()) {
    if (!fs.existsSync(target)) {
      fs.mkdirSync(target);
    }
    let files = fs.readdirSync(source);
    files.forEach((file) => {
      recursiveCopySync(path.join(source, file), path.join(target, file));
    });
  } else {
    if (fs.existsSync(source)) {
      fs.writeFileSync(target, fs.readFileSync(source));
    }
  }
}

Mallikarjun M,谢谢!

fs-extra做了这件事,如果你不提供一个回调,它甚至可以返回一个承诺!:)

const path = require('path')
const fs = require('fs-extra')

let source = path.resolve( __dirname, 'folderA')
let destination = path.resolve( __dirname, 'folderB')

fs.copy(source, destination)
  .then(() => console.log('Copy completed!'))
  .catch( err => {
    console.log('An error occurred while copying the folder.')
    return console.error(err)
  })

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);
    }
  });
}