node.js Express.js - 显示文件/目录列表的任何方式？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/6294186/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Express.js - any way to display a file/dir listing?
提问by balupton
With Express.jsis there a way to display a file/dir listing like apache does when you access the URL of a directory which doesn't have a index file - so it displays a listing of all that directories contents?
随着Express.js是有显示的文件/目录列表就像当你访问它没有一个索引文件目录的URL Apache没有办法-所以它显示的所有目录中的内容的列表？
Is there an extension or package that does this which I don't know of? Or will I have to code this myself?
是否有我不知道的扩展或包可以做到这一点？还是我必须自己编码？
Cheers guys, you rock! :)
干杯伙计们，你摇滚！:)
回答by yonran
There's a brand new default Connect middleware named directory(source) for directory listings. It has a lot of style and has a client-side search box.
有一个名为directory( source) 的全新默认Connect 中间件用于目录列表。它有很多风格，并有一个客户端搜索框。
var express = require('express')
, app = express.createServer();
app.configure(function() {
var hourMs = 1000*60*60;
app.use(express.static(__dirname + '/public', { maxAge: hourMs }));
app.use(express.directory(__dirname + '/public'));
app.use(express.errorHandler());
});
app.listen(8080);
回答by jbll
As of Express 4.x, the directory middleware is no longer bundled with express. You'll want to download the npm module serve-index.
从 Express 4.x 开始，目录中间件不再与 express 捆绑在一起。您需要下载 npm 模块serve-index。
Then, for example, to display the file/dir listings in a directory at the root of the app called videoswould look like:
然后，例如，要在被调用的应用程序根目录下的目录中显示文件/目录列表，videos将如下所示：
var serveIndex = require('serve-index');
app.use(express.static(__dirname + "/"))
app.use('/videos', serveIndex(__dirname + '/videos'));
回答by Talespin_Kit
The following code will serve both directory and files
以下代码将同时提供目录和文件
var serveIndex = require('serve-index');
app.use('/p', serveIndex(path.join(__dirname, 'public')));
app.use('/p', express.static(path.join(__dirname, 'public')));
回答by Xin
This will do the work for you: (new version of express requires separate middleware). E.g. you put your files under folder 'files' and you want the url to be '/public'
这将为您完成工作：（新版本的 express 需要单独的中间件）。例如，您将文件放在文件夹“files”下，并且希望 url 为“/public”
var express = require('express');
var serveIndex = require('serve-index');
var app = express();
app.use('/public', serveIndex('files')); // shows you the file list
app.use('/public', express.static('files')); // serve the actual files
回答by Matteljay
Built-in NodeJS module fsgives a lot of fine-grained options
内置的 NodeJS 模块fs提供了很多细粒度的选项
const fs = require('fs')
router.get('*', (req, res) => {
const fullPath = process.cwd() + req.path //(not __dirname)
const dir = fs.opendirSync(fullPath)
let entity
let listing = []
while((entity = dir.readSync()) !== null) {
if(entity.isFile()) {
listing.push({ type: 'f', name: entity.name })
} else if(entity.isDirectory()) {
listing.push({ type: 'd', name: entity.name })
}
}
dir.closeSync()
res.send(listing)
})
Please make sure to read up on path-traversal security vulnerabilities.
请务必阅读路径遍历安全漏洞。
