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
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));
}
其他回答
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。
这是我解决这个问题的方法,没有任何额外的模块。只使用内置的fs和path模块。
注意:这使用fs的读/写函数,所以它不复制任何元数据(创建时间等)。从Node.js 8.5开始,有一个copyFileSync函数可用,它调用操作系统复制函数,因此也复制元数据。我还没有测试它们,但它应该可以替换它们。(见https://nodejs.org/api/fs.html fs_fs_copyfilesync_src_dest_flags)
var fs = require('fs');
var path = require('path');
function copyFileSync( source, target ) {
var targetFile = target;
// If target is a directory, a new file with the same name will be created
if ( fs.existsSync( target ) ) {
if ( fs.lstatSync( target ).isDirectory() ) {
targetFile = path.join( target, path.basename( source ) );
}
}
fs.writeFileSync(targetFile, fs.readFileSync(source));
}
function copyFolderRecursiveSync( source, target ) {
var files = [];
// Check if folder needs to be created or integrated
var targetFolder = path.join( target, path.basename( source ) );
if ( !fs.existsSync( targetFolder ) ) {
fs.mkdirSync( targetFolder );
}
// Copy
if ( fs.lstatSync( source ).isDirectory() ) {
files = fs.readdirSync( source );
files.forEach( function ( file ) {
var curSource = path.join( source, file );
if ( fs.lstatSync( curSource ).isDirectory() ) {
copyFolderRecursiveSync( curSource, targetFolder );
} else {
copyFileSync( curSource, targetFolder );
}
} );
}
}
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
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。
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)