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

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)
  })

其他回答

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);
        }
    });
}

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。

下面是一个递归复制目录及其内容到另一个目录的函数:

const fs = require("fs")
const path = require("path")

/**
 * Look ma, it's cp -R.
 * @param {string} src  The path to the thing to copy.
 * @param {string} dest The path to the new copy.
 */
var copyRecursiveSync = function(src, dest) {
  var exists = fs.existsSync(src);
  var stats = exists && fs.statSync(src);
  var isDirectory = exists && stats.isDirectory();
  if (isDirectory) {
    fs.mkdirSync(dest);
    fs.readdirSync(src).forEach(function(childItemName) {
      copyRecursiveSync(path.join(src, childItemName),
                        path.join(dest, childItemName));
    });
  } else {
    fs.copyFileSync(src, dest);
  }
};

我知道这里已经有很多答案了,但是没有一个答案是简单的。

关于fs-exra官方文档,您可以非常轻松地完成。

const fs = require('fs-extra')

// Copy file
fs.copySync('/tmp/myfile', '/tmp/mynewfile')

// Copy directory, even if it has subdirectories or files
fs.copySync('/tmp/mydir', '/tmp/mynewdir')

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));
}