是否有一种方法可以使用Java SE API在Java中创建一个非常基本的HTTP服务器(只支持GET/POST),而不需要编写代码手动解析HTTP请求和手动格式化HTTP响应?Java SE API在HttpURLConnection中很好地封装了HTTP客户机功能,但是是否有类似的HTTP服务器功能呢?

需要明确的是,我在网上看到的许多ServerSocket示例的问题是,它们自己进行请求解析/响应格式化和错误处理,这很乏味,容易出错,而且不太全面,出于这些原因,我试图避免使用它。


当前回答

这个代码比我们的代码更好,你只需要添加2个库:javax. servlet .jar和org.mortbay.jetty.jar。

类码头:

package jetty;

import java.util.logging.Level;
import java.util.logging.Logger;
import org.mortbay.http.SocketListener;
import org.mortbay.jetty.Server;
import org.mortbay.jetty.servlet.ServletHttpContext;

public class Jetty {

    public static void main(String[] args) {
        try {
            Server server = new Server();
            SocketListener listener = new SocketListener();      

            System.out.println("Max Thread :" + listener.getMaxThreads() + " Min Thread :" + listener.getMinThreads());

            listener.setHost("localhost");
            listener.setPort(8070);
            listener.setMinThreads(5);
            listener.setMaxThreads(250);
            server.addListener(listener);            

            ServletHttpContext context = (ServletHttpContext) server.getContext("/");
            context.addServlet("/MO", "jetty.HelloWorldServlet");

            server.start();
            server.join();

        /*//We will create our server running at http://localhost:8070
        Server server = new Server();
        server.addListener(":8070");

        //We will deploy our servlet to the server at the path '/'
        //it will be available at http://localhost:8070
        ServletHttpContext context = (ServletHttpContext) server.getContext("/");
        context.addServlet("/MO", "jetty.HelloWorldServlet");

        server.start();
        */

        } catch (Exception ex) {
            Logger.getLogger(Jetty.class.getName()).log(Level.SEVERE, null, ex);
        }

    }
} 

Servlet类:

package jetty;

import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class HelloWorldServlet extends HttpServlet
{
    @Override
    protected void doGet(HttpServletRequest httpServletRequest, HttpServletResponse httpServletResponse) throws ServletException, IOException
    {
        String appid = httpServletRequest.getParameter("appid");
        String conta = httpServletRequest.getParameter("conta");

        System.out.println("Appid : "+appid);
        System.out.println("Conta : "+conta);

        httpServletResponse.setContentType("text/plain");
        PrintWriter out = httpServletResponse.getWriter();
        out.println("Hello World!");
        out.close();
    }
}

其他回答

httpserver解决方案不能跨jre移植。最好使用javax.xml.ws中的官方webservices API来引导一个最小的HTTP服务器…

import java.io._
import javax.xml.ws._
import javax.xml.ws.http._
import javax.xml.transform._
import javax.xml.transform.stream._

@WebServiceProvider
@ServiceMode(value=Service.Mode.PAYLOAD) 
class P extends Provider[Source] {
  def invoke(source: Source) = new StreamSource( new StringReader("<p>Hello There!</p>"));
}

val address = "http://127.0.0.1:8080/"
Endpoint.create(HTTPBinding.HTTP_BINDING, new P()).publish(address)

println("Service running at "+address)
println("Type [CTRL]+[C] to quit!")

Thread.sleep(Long.MaxValue)

编辑:这实际上是工作!上面的代码看起来像Groovy之类的。以下是我测试的Java翻译:

import java.io.*;
import javax.xml.ws.*;
import javax.xml.ws.http.*;
import javax.xml.transform.*;
import javax.xml.transform.stream.*;

@WebServiceProvider
@ServiceMode(value = Service.Mode.PAYLOAD)
public class Server implements Provider<Source> {

    public Source invoke(Source request) {
        return  new StreamSource(new StringReader("<p>Hello There!</p>"));
    }

    public static void main(String[] args) throws InterruptedException {

        String address = "http://127.0.0.1:8080/";
        Endpoint.create(HTTPBinding.HTTP_BINDING, new Server()).publish(address);

        System.out.println("Service running at " + address);
        System.out.println("Type [CTRL]+[C] to quit!");

        Thread.sleep(Long.MAX_VALUE);
    }
}

检查拍摄。查看https://github.com/yegor256/takes获取快速信息

从Java SE 6开始,在Sun Oracle JRE中有一个内置的HTTP服务器。Java 9模块名称为jdk.httpserver。httpserver包摘要概述了涉及的类并包含示例。

这里有一个从他们的文档复制粘贴的启动示例。你可以复制,粘贴,然后在Java 6+上运行。 (尽管如此,所有试图编辑它的人,因为它是一段丑陋的代码,请不要,这是一个复制粘贴,不是我的,此外,你不应该编辑引文,除非它们在原始来源中发生了变化)

package com.stackoverflow.q3732109; import java.io.IOException; import java.io.OutputStream; import java.net.InetSocketAddress; import com.sun.net.httpserver.HttpExchange; import com.sun.net.httpserver.HttpHandler; import com.sun.net.httpserver.HttpServer; public class Test { public static void main(String[] args) throws Exception { HttpServer server = HttpServer.create(new InetSocketAddress(8000), 0); server.createContext("/test", new MyHandler()); server.setExecutor(null); // creates a default executor server.start(); } static class MyHandler implements HttpHandler { @Override public void handle(HttpExchange t) throws IOException { String response = "This is the response"; t.sendResponseHeaders(200, response.length()); OutputStream os = t.getResponseBody(); os.write(response.getBytes()); os.close(); } } }

应该注意的是,他们示例中的response.length()部分是坏的,它应该是response.getBytes().length。即使这样,getBytes()方法也必须显式地指定在响应头中指定的字符集。唉,尽管对初学者有误导,但毕竟这只是一个基本的启动示例。

执行它并访问http://localhost:8000/test,你将看到以下响应:

这是反应


As to using com.sun.* classes, do note that this is, in contrary to what some developers think, absolutely not forbidden by the well known FAQ Why Developers Should Not Write Programs That Call 'sun' Packages. That FAQ concerns the sun.* package (such as sun.misc.BASE64Encoder) for internal usage by the Oracle JRE (which would thus kill your application when you run it on a different JRE), not the com.sun.* package. Sun/Oracle also just develop software on top of the Java SE API themselves like as every other company such as Apache and so on. Moreover, this specific HttpServer must be present in every JDK so there is absolutely no means of "portability" issue like as would happen with sun.* package. Using com.sun.* classes is only discouraged (but not forbidden) when it concerns an implementation of a certain Java API, such as GlassFish (Java EE impl), Mojarra (JSF impl), Jersey (JAX-RS impl), etc.

检出简单。它是一个非常简单的嵌入式服务器,内置了对各种操作的支持。我特别喜欢它的线程模型..

神奇的!

从Java 18开始,你可以用Java标准库创建简单的web服务器:

class Main {
    public static void main(String[] args) {
        var port = 8000;
        var rootDirectory = Path.of("C:/Users/Mahozad/Desktop/");
        var outputLevel = OutputLevel.VERBOSE;
        var server = SimpleFileServer.createFileServer(
                new InetSocketAddress(port),
                rootDirectory,
                outputLevel
        );
        server.start();
    }
}

默认情况下,这将显示指定根目录的目录列表。您可以将index.html文件(以及CSS和JS文件等其他资产)放在该目录中来显示它们。

示例(我把这些放在桌面上,上面指定为我的根目录):

index . html:

<!DOCTYPE html>
<html>
<head>
  <meta charset="UTF-8">
  <title>Java 18 Simple Web Server</title>
  <link rel="stylesheet" href="styles.css">
  <style>h1 { color: blue; }</style>
  <script src="scripts.js" defer>
    let element = document.getElementsByTagName("h1")[0];
    element.style.fontSize = "48px";
  </script>
</head>
<body>
  <h1>I'm <i>index.html</i> in the root directory.</h1>
</body>
</html>

旁注

对于Java标准库HTTP客户端,请参阅Java 11新HTTP客户端API。