使用 Node.js 将文件系统中的目录结构转换为 JSON

我有一个这样的文件结构:

root
|_ fruits
|___ apple
|______images
|________ apple001.jpg
|________ apple002.jpg
|_ animals
|___ cat
|______images
|________ cat001.jpg
|________ cat002.jpg

我想使用 Javascript 和 Node.js，监听这个根目录和所有子目录，并创建一个镜像这个目录结构的 JSON，每个节点包含类型、名称、路径和子节点:

data = [
{
type: "folder",
name: "animals",
path: "/animals",
children: [
{
type: "folder",
name: "cat",
path: "/animals/cat",
children: [
{
type: "folder",
name: "images",
path: "/animals/cat/images",
children: [
{
type: "file",
name: "cat001.jpg",
path: "/animals/cat/images/cat001.jpg"
}, {
type: "file",
name: "cat001.jpg",
path: "/animals/cat/images/cat002.jpg"
}
]
}
]
}
]
}
];

这里有一个咖啡脚本 JSON:

data =
[
type: "folder"
name: "animals"
path: "/animals"
children  :
[
type: "folder"
name: "cat"
path: "/animals/cat"
children:
[
type: "folder"
name: "images"
path: "/animals/cat/images"
children:
[
type: "file"
name: "cat001.jpg"
path: "/animals/cat/images/cat001.jpg"
,
type: "file"
name: "cat001.jpg"
path: "/animals/cat/images/cat002.jpg"
]
]
]
]

如何在 django 视图中获得这种 json 数据格式? (python)

90174 次浏览
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最佳答案

这是一个草图。错误处理留给读者作为练习。

var fs = require('fs'),
path = require('path')


function dirTree(filename) {
var stats = fs.lstatSync(filename),
info = {
path: filename,
name: path.basename(filename)
};


if (stats.isDirectory()) {
info.type = "folder";
info.children = fs.readdirSync(filename).map(function(child) {
return dirTree(filename + '/' + child);
});
} else {
// Assuming it's a file. In real life it could be a symlink or
// something else!
info.type = "file";
}


return info;
}


if (module.parent == undefined) {
// node dirTree.js ~/foo/bar
var util = require('util');
console.log(util.inspect(dirTree(process.argv[2]), false, null));
}
小开

您可以使用这个项目中的代码，但是您应该根据自己的需要调整代码:

Https://github.com/nhq/node-fileutils/blob/master/src/file-utils.js#l511-l593

来自:

a
|- b
|  |- c
|  |  |- c1.txt
|  |
|  |- b1.txt
|  |- b2.txt
|
|- d
|  |
|
|- a1.txt
|- a2.txt

致:

{
b: {
"b1.txt": "a/b/b1.txt",
"b2.txt": "a/b/b2.txt",
c: {
"c1.txt": "a/b/c/c1.txt"
}
},
d: {},
"a2.txt": "a/a2.txt",
"a1.txt": "a/a1.txt"
}

做法:

new File ("a").list (function (error, files){
//files...
});
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我的 CS 示例(w/Express)基于 Miika 的解决方案:

fs = require 'fs' #file system module
path = require 'path' # file path module


# returns json tree of directory structure
tree = (root) ->
# clean trailing '/'(s)
root = root.replace /\/+$/ , ""
# extract tree ring if root exists
if fs.existsSync root
ring = fs.lstatSync root
else
return 'error: root does not exist'
# type agnostic info
info =
path: root
name: path.basename(root)
# dir
if ring.isDirectory()
info.type = 'folder'
# execute for each child and call tree recursively
info.children = fs.readdirSync(root) .map (child) ->
tree root + '/' + child
# file
else if ring.isFile()
info.type = 'file'
# link
else if ring.isSymbolicLink()
info.type = 'link'
# other
else
info.type = 'unknown'
# return tree
info


# error handling
handle = (e) ->
return 'uncaught exception...'


exports.index = (req, res) ->
try
res.send tree './test/'
catch e
res.send handle e
小开

可以接受的答案是有效的，但它是 同步，这会严重影响您的性能，特别是对于大型目录树。
我强烈建议您使用下面的 异步的解决方案，它既快又非阻塞。
基于并行解 给你。

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


var diretoryTreeToObj = function(dir, done) {
var results = [];


fs.readdir(dir, function(err, list) {
if (err)
return done(err);


var pending = list.length;


if (!pending)
return done(null, {name: path.basename(dir), type: 'folder', children: results});


list.forEach(function(file) {
file = path.resolve(dir, file);
fs.stat(file, function(err, stat) {
if (stat && stat.isDirectory()) {
diretoryTreeToObj(file, function(err, res) {
results.push({
name: path.basename(file),
type: 'folder',
children: res
});
if (!--pending)
done(null, results);
});
}
else {
results.push({
type: 'file',
name: path.basename(file)
});
if (!--pending)
done(null, results);
}
});
});
});
};

示例用法:

var dirTree = ('/path/to/dir');


diretoryTreeToObj(dirTree, function(err, res){
if(err)
console.error(err);


console.log(JSON.stringify(res));
});
小开

有一个 NPM 模块

Https://www.npmjs.com/package/directory-tree

创建表示目录树的对象。

来自:

photos
├── summer
│   └── june
│       └── windsurf.jpg
└── winter
└── january
├── ski.png
└── snowboard.jpg

致:

{
"path": "",
"name": "photos",
"type": "directory",
"children": [
{
"path": "summer",
"name": "summer",
"type": "directory",
"children": [
{
"path": "summer/june",
"name": "june",
"type": "directory",
"children": [
{
"path": "summer/june/windsurf.jpg",
"name": "windsurf.jpg",
"type": "file"
}
]
}
]
},
{
"path": "winter",
"name": "winter",
"type": "directory",
"children": [
{
"path": "winter/january",
"name": "january",
"type": "directory",
"children": [
{
"path": "winter/january/ski.png",
"name": "ski.png",
"type": "file"
},
{
"path": "winter/january/snowboard.jpg",
"name": "snowboard.jpg",
"type": "file"
}
]
}
]
}
]
}

用法

var tree = directoryTree('/some/path');

你也可以通过扩展来过滤:

var filteredTree = directoryTree('/some/path', ['.jpg', '.png']);
小开

我在这个例子中使用了‘ walk’lib，它获取你的根路径，递归遍历文件和目录，并发出一个目录/文件事件，其中包含你需要的所有节点信息, 检查一下这个实现—— > 

const walk = require('walk');


class FsTree {


constructor(){


}


/**
* @param rootPath
* @returns {Promise}
*/
getFileSysTree(rootPath){
return new Promise((resolve, reject)=>{


const root = rootPath || __dirname; // if there's no rootPath use exec location
const tree = [];
const nodesMap = {};
const walker  = walk.walk(root, { followLinks: false}); // filter doesn't work well


function addNode(node, path){
if ( node.name.indexOf('.') === 0 || path.indexOf('/.') >= 0){ // ignore hidden files
return;
}
var relativePath = path.replace(root,'');


node.path = relativePath + '/' + node.name;
nodesMap[node.path] = node;


if ( relativePath.length === 0 ){ //is root
tree.push(node);
return;
}
node.parentPath = node.path.substring(0,node.path.lastIndexOf('/'));
const parent = nodesMap[node.parentPath];
parent.children.push(node);


}


walker.on('directory', (path, stats, next)=>{
addNode({ name: stats.name, type:'dir',children:[]}, path);
next();
});


walker.on('file', (path,stats,next)=>{
addNode({name:stats.name, type:'file'},path);
next();
});


walker.on('end',()=>{
resolve(tree);
});


walker.on('errors',  (root, nodeStatsArray, next) => {
reject(nodeStatsArray);
next();
});
});


}
}




const fsTreeFetcher = new FsTree();


fsTreeFetcher.getFileSysTree(__dirname).then((result)=>{
console.log(result);
});
小开

下面是一个异步解决方案:

 function list(dir) {
const walk = entry => {
return new Promise((resolve, reject) => {
fs.exists(entry, exists => {
if (!exists) {
return resolve({});
}
return resolve(new Promise((resolve, reject) => {
fs.lstat(entry, (err, stats) => {
if (err) {
return reject(err);
}
if (!stats.isDirectory()) {
return resolve({
// path: entry,
// type: 'file',
name: path.basename(entry),
time: stats.mtime,
size: stats.size
});
}
resolve(new Promise((resolve, reject) => {
fs.readdir(entry, (err, files) => {
if (err) {
return reject(err);
}
Promise.all(files.map(child => walk(path.join(entry, child)))).then(children => {
resolve({
// path: entry,
// type: 'folder',
name: path.basename(entry),
time: stats.mtime,
entries: children
});
}).catch(err => {
reject(err);
});
});
}));
});
}));
});
});
}


return walk(dir);
}

请注意，当目录不存在时，将返回一个空结果，而不是抛出一个错误。

下面是一个样本结果:

{
"name": "root",
"time": "2017-05-09T07:46:26.740Z",
"entries": [
{
"name": "book.txt",
"time": "2017-05-09T07:24:18.673Z",
"size": 0
},
{
"name": "cheatsheet-a5.pdf",
"time": "2017-05-09T07:24:18.674Z",
"size": 262380
},
{
"name": "docs",
"time": "2017-05-09T07:47:39.507Z",
"entries": [
{
"name": "README.md",
"time": "2017-05-08T10:02:09.651Z",
"size": 19229
}
]
}
]
}

就是:

root
|__ book.txt
|__ cheatsheet-a5.pdf
|__ docs
|__ README.md
小开

综合 肖恩 · C。的答案。

我非常喜欢它，但是使用异步等待使它更具可读性。

import fs from 'fs';
import {
lstat,
readdir,
access,
} from 'fs/promises';
import path from 'path';


async function existsAsync(file) {
try {
await access(file, fs.constants.F_OK);
return true;
} catch (e) {
return false;
}
}


async function listFileTreeRecursive(dir) {
const recurse = async (entry) => {
if (!(await existsAsync(entry))) {
return {};
}


const stats = await lstat(entry);
if (!stats.isDirectory()) {
return {
name: path.basename(entry),
time: stats.mtime,
size: stats.size,
};
}


const files = await readdir(entry);
const childEntries = await Promise.all(
files.map((child) => recurse(path.join(entry, child))),
);
return {
name: path.basename(entry),
time: stats.mtime,
entries: childEntries,
};
};


return recurse(dir);
}

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