java - 如何让 HttpServer 并行创建多个 HttpHandlers？

Question

我正在使用 com.sun.net.httpserver.HttpServer 创建一个小容器来测试服务器代码位，但我无法让它使用多个线程来处理请求。

我调用 java.util.concurrent.Executors.newFixedThreadPool(20) 创建一个具有 20 个线程的 java.util.concurrent.ThreadPoolExecutor。然后，我在 HttpServer 上设置了这个 Executor。使用 Jmeter，我触发了 20 个客户端线程来发送一个请求，该请求将被路由到服务器中唯一的 HttpHandler 实现。该处理程序执行 System.out.println(this) 并且我看到了以下输出:

Started TestServer at port 8800
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa
http.TestHandler@30eb9dfa

我想我会看到这里使用了 20 个(或将近 20 个)不同的线程。这是代码。

package http;

import java.io.IOException;
import java.io.OutputStream;
import java.net.InetSocketAddress;
import java.util.concurrent.Executors;
import java.util.concurrent.ExecutorService;

import com.sun.net.httpserver.HttpExchange;
import com.sun.net.httpserver.HttpHandler;
import com.sun.net.httpserver.HttpServer;

public class TestServer implements Runnable {

    private final static int PORT    = Integer.getInteger("test.port", 8800); 
    private static TestServer serverInstance;
    private HttpServer        httpServer;
    private ExecutorService   executor;

    @Override
    public void run() {
        try {
            executor = Executors.newFixedThreadPool(20);

            httpServer = HttpServer.create(new InetSocketAddress(PORT), 0);
            httpServer.createContext("/test", new TestHandler());
            httpServer.setExecutor(executor);
            httpServer.start();
            System.out.println("Started TestServer at port " + PORT);

            // Wait here until notified of shutdown.
            synchronized (this) {
                try {
                    this.wait();
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
        } catch (Throwable t) {
            t.printStackTrace();
        }
    }

    static void shutdown() {

        try { 
            System.out.println("Shutting down TestServer.");            
            serverInstance.httpServer.stop(0);

        } catch (Exception e) {
            e.printStackTrace();
        }

        synchronized (serverInstance) {
            serverInstance.notifyAll();
        }

    }

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

        serverInstance = new TestServer();

        Thread serverThread = new Thread(serverInstance);
        serverThread.start();

        Runtime.getRuntime().addShutdownHook(new OnShutdown());

        try {
            serverThread.join();
        } catch (Exception e) { }
    }

}

/* Responds to the /test URI. */
class TestHandler implements HttpHandler {

    boolean debug = Boolean.getBoolean("test.debug");

    public void handle(HttpExchange exchange) throws IOException {

        System.out.println(this);  // ALWAYS SAME THREAD!

        String response = "RESPONSE AT " + System.currentTimeMillis();

        exchange.sendResponseHeaders(200, response.length());
        OutputStream os = exchange.getResponseBody();
        os.write(response.getBytes());
        os.flush();
        os.close();
    }
}

/* Responds to a JVM shutdown by stopping the server. */
class OnShutdown extends Thread {
    public void run() {
        TestServer.shutdown();
    }
}

我希望 HttpServer 并行创建多个 TestHandler 来为多个并发请求提供服务。我在这里缺少什么？

(顺便说一句，这与 Can I make a Java HttpServer threaded/process requests in parallel? 非常相似，尽管答案是使用 Executor，我已经在这样做了。谢谢。)

Answer 1

同一可运行实例可以在不同线程中运行多次。参见 Initializing two threads with the same instance of a runnable了解更多信息。

您在示例中打印的是 HttpHandler 信息，但没有关于哪个线程正在运行的信息。并且该信息不会改变，因为服务器始终为所有线程重用相同的对象。

如果你想打印线程id，你可以使用:

long threadId = Thread.currentThread().getId();
System.out.println(threadId);

threadId 应按预期更改。