我想运行一个非常简单的HTTP服务器。对example.com的每个GET请求都应该得到index.html,但作为一个常规的HTML页面(即,与阅读普通网页时的体验相同)。

使用下面的代码,我可以读取index.html的内容。我如何服务index.html作为一个普通的网页?

var http = require('http');
var fs = require('fs');
var index = fs.readFileSync('index.html');

http.createServer(function (req, res) {
  res.writeHead(200, {'Content-Type': 'text/plain'});
  res.end(index);
}).listen(9615);

下面的一个建议很复杂,需要我为我想使用的每个资源(CSS、JavaScript、图像)文件写一个get行。

我如何能提供一个单一的HTML页面与一些图像,CSS和JavaScript?


当前回答

你不需要快递。你不需要联系。Node.js执行http native。你所需要做的就是根据请求返回一个文件:

var http = require('http')
var url = require('url')
var fs = require('fs')

http.createServer(function (request, response) {
    var requestUrl = url.parse(request.url)    
    response.writeHead(200)
    fs.createReadStream(requestUrl.pathname).pipe(response)  // do NOT use fs's sync methods ANYWHERE on production (e.g readFileSync) 
}).listen(9615)    

一个更完整的例子,确保请求不能访问基目录下的文件,并进行适当的错误处理:

var http = require('http')
var url = require('url')
var fs = require('fs')
var path = require('path')
var baseDirectory = __dirname   // or whatever base directory you want

var port = 9615

http.createServer(function (request, response) {
    try {
        var requestUrl = url.parse(request.url)

        // need to use path.normalize so people can't access directories underneath baseDirectory
        var fsPath = baseDirectory+path.normalize(requestUrl.pathname)

        var fileStream = fs.createReadStream(fsPath)
        fileStream.pipe(response)
        fileStream.on('open', function() {
             response.writeHead(200)
        })
        fileStream.on('error',function(e) {
             response.writeHead(404)     // assume the file doesn't exist
             response.end()
        })
   } catch(e) {
        response.writeHead(500)
        response.end()     // end the response so browsers don't hang
        console.log(e.stack)
   }
}).listen(port)

console.log("listening on port "+port)

其他回答

编辑:

Node.js示例应用Node Chat拥有你想要的功能。 在它的readme。textfile中 3.步伐就是你要找的。

step1 创建一个在端口8002上响应hello world的服务器 步骤2 创建一个index.html并服务它 步骤3 介绍util.js 更改逻辑,以便为任何静态文件提供服务 如果找不到文件,显示404 第四 添加jquery-1.4.2.js 添加client.js 将index.html更改为提示用户昵称

这是server.js

这是util.js

我使用下面的代码来启动一个简单的web服务器,如果Url中没有提到文件,它会渲染默认的html文件。

var http = require('http'),
fs = require('fs'),
url = require('url'),
rootFolder = '/views/',
defaultFileName = '/views/5 Tips on improving Programming Logic   Geek Files.htm';


http.createServer(function(req, res){

    var fileName = url.parse(req.url).pathname;
    // If no file name in Url, use default file name
    fileName = (fileName == "/") ? defaultFileName : rootFolder + fileName;

    fs.readFile(__dirname + decodeURIComponent(fileName), 'binary',function(err, content){
        if (content != null && content != '' ){
            res.writeHead(200,{'Content-Length':content.length});
            res.write(content);
        }
        res.end();
    });

}).listen(8800);

它将呈现所有的js, css和图像文件,以及所有的html内容。

同意“没有内容类型比错误的内容类型更好”的说法

你不需要快递。你不需要联系。Node.js执行http native。你所需要做的就是根据请求返回一个文件:

var http = require('http')
var url = require('url')
var fs = require('fs')

http.createServer(function (request, response) {
    var requestUrl = url.parse(request.url)    
    response.writeHead(200)
    fs.createReadStream(requestUrl.pathname).pipe(response)  // do NOT use fs's sync methods ANYWHERE on production (e.g readFileSync) 
}).listen(9615)    

一个更完整的例子,确保请求不能访问基目录下的文件,并进行适当的错误处理:

var http = require('http')
var url = require('url')
var fs = require('fs')
var path = require('path')
var baseDirectory = __dirname   // or whatever base directory you want

var port = 9615

http.createServer(function (request, response) {
    try {
        var requestUrl = url.parse(request.url)

        // need to use path.normalize so people can't access directories underneath baseDirectory
        var fsPath = baseDirectory+path.normalize(requestUrl.pathname)

        var fileStream = fs.createReadStream(fsPath)
        fileStream.pipe(response)
        fileStream.on('open', function() {
             response.writeHead(200)
        })
        fileStream.on('error',function(e) {
             response.writeHead(404)     // assume the file doesn't exist
             response.end()
        })
   } catch(e) {
        response.writeHead(500)
        response.end()     // end the response so browsers don't hang
        console.log(e.stack)
   }
}).listen(port)

console.log("listening on port "+port)

这是一个最快的解决方案,我用快速看到网页

sudo npm install ripple-emulator -g

从那时起,只需输入html文件的目录并运行

ripple emulate

然后将设备更改为Nexus 7横屏。

这很容易,因为今天有大量的图书馆。这里的答案是功能性的。如果你想要另一个版本开始更快和简单

当然,首先要安装node.js。后:

> # module with zero dependencies
> npm install -g @kawix/core@latest 
> # change /path/to/static with your folder or empty for current
> kwcore "https://raw.githubusercontent.com/voxsoftware/kawix-core/master/example/npmrequire/express-static.js" /path/to/static

这里是“https://raw.githubusercontent.com/voxsoftware/kawix-core/master/example/npmrequire/express-static.js”的内容(你不需要下载,我贴出来是为了了解后面的工作原理)

// you can use like this:
// kwcore "https://raw.githubusercontent.com/voxsoftware/kawix-core/master/example/npmrequire/express.js" /path/to/static
// kwcore "https://raw.githubusercontent.com/voxsoftware/kawix-core/master/example/npmrequire/express.js" 

// this will download the npm module and make a local cache
import express from 'npm://express@^4.16.4'
import Path from 'path'

var folder= process.argv[2] || "."
folder= Path.resolve(process.cwd(), folder)
console.log("Using folder as public: " + folder)

var app = express() 
app.use(express.static(folder)) 
app.listen(8181)
console.log("Listening on 8181")