用node.js写一个简单的服务器，只使用了20行左右的代码，支持根据视频名字点播，视频文件和node.js文件放在同一目录下；
服务器代码如下：
var express = require('express');
var pg = require('pg');
var app = express();
var fs = require('fs');
app.get('/video', function (req, res) {
var time = new Date();
var videoName = req.query.name;
console.log("-------点击查询下载" + time.getFullYear() + "/" + time.getMonth() + "/" + time.getDate() + "/" + time.getHours() + "/" + time.getMinutes() + "/" + time.getSeconds() + "-------");
res.writeHead(200, {'Content-Type': 'video/mp4'});
var rs = fs.createReadStream(videoName + '.mp4');
rs.pipe(res);
rs.on('end', function () {
res.end();
console.log('end call');
});
});
var server = app.listen(8081, function () {
var host = server.address().address;
var port = server.address().port;
console.log("应用实例，访问地址为 http://%s:%s", host, port);
});测试播放html代码为
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>视频播放</title>
</head>
<body>
<video width="320" height="240" controls="controls">
<source src="http://localhost:8081/video?name=123" type="video/mp4">
</video>
</body>
</html>结果为：
视频测试文件为123.mp4
