SignalR java无法连接到服务器

Question

我将我的java应用程序连接到SignalR Server时遇到问题.
服务器非常简单,可以在这里找到:https:
//code.msdn.microsoft.com/windowsdesktop/Using-SignalR-in-WinForms-f1ec847b

我可以连接Web客户端(javascript)和Windows客户端(C#)但我的java客户端有问题.(https://github.com/SignalR/java-client)

到目前为止,这是我的代码:

package javaapplication2;

import java.io.IOException;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.TimeoutException;
import microsoft.aspnet.signalr.client.SignalRFuture;
import microsoft.aspnet.signalr.client.hubs.HubConnection;
import microsoft.aspnet.signalr.client.hubs.HubProxy;

public class JavaApplication2 {
       public static void main(String[] args) throws IOException, InterruptedException, ExecutionException, TimeoutException 
       {
            String ServerURI = "http://localhost:8080/signalr";
            HubConnection Connection = new HubConnection(ServerURI);
            HubProxy HubProxy = Connection.createHubProxy("MyHub");
            HubProxy.on("AddMessage", () -> { System.out.println("Some message");        });

            Connection.error(new ErrorCallback() {
                @Override
                public void onError(Throwable error) {
                    error.printStackTrace(); //<==SocketException
                }
            });

            SignalRFuture<Void> con  =Connection.start();
            con.get(); 
       }
}

更新

当我运行它时,我得到一个"ExecutionException:java.net.SocketException:Connection reset"

Exception in thread "main" java.util.concurrent.ExecutionException: java.net.SocketException: Connection reset
at microsoft.aspnet.signalr.client.SignalRFuture.get(SignalRFuture.java:112)
at microsoft.aspnet.signalr.client.SignalRFuture.get(SignalRFuture.java:102)
at javaapplication2.JavaApplication2.main(JavaApplication2.java:27)
Caused by: java.net.SocketException: Connection reset
at java.net.SocketInputStream.read(SocketInputStream.java:209)
at java.net.SocketInputStream.read(SocketInputStream.java:141)
at java.io.BufferedInputStream.fill(BufferedInputStream.java:246)
at java.io.BufferedInputStream.read1(BufferedInputStream.java:286)
at java.io.BufferedInputStream.read(BufferedInputStream.java:345)
at sun.net.www.http.HttpClient.parseHTTPHeader(HttpClient.java:704)
at sun.net.www.http.HttpClient.parseHTTP(HttpClient.java:647)
at sun.net.www.http.HttpClient.parseHTTP(HttpClient.java:675)
at sun.net.www.protocol.http.HttpURLConnection.getInputStream0(HttpURLConnection.java:1535)
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1440)
at java.net.HttpURLConnection.getResponseCode(HttpURLConnection.java:480)
at microsoft.aspnet.signalr.client.http.java.NetworkRunnable.run(NetworkRunnable.java:72)
at java.lang.Thread.run(Thread.java:745)

- 如果我将"localhost"更改为不存在的东西(例如locahostX),我会得到一个java.net.UnknownHostException

- 如果将"localhost"更改为我的IP我没有事件获得异常...

- 所有其他应用程序都适用于(localhost或IP)

起初我以为这是一个防火墙问题,但不是......

显然我错过了什么......有什么想法吗？

谢谢

Answer 1

事实证明，我必须使用 start 的重载，该重载将 ClientTransport 对象作为参数

public SignalRFuture<Void> start(ClientTransport transport)

如果有人对无参数启动方法失败的原因有解释，请将其作为答案发布，我会将其标记为解决方案。

这是一个有效的完整示例：

package javaapplication2;

import java.io.IOException;
import java.util.Scanner;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.TimeoutException;
import microsoft.aspnet.signalr.client.SignalRFuture;
import microsoft.aspnet.signalr.client.hubs.HubConnection;
import microsoft.aspnet.signalr.client.hubs.HubProxy;
import microsoft.aspnet.signalr.client.transport.ServerSentEventsTransport;
import microsoft.aspnet.signalr.client.hubs.SubscriptionHandler2;

public class JavaApplication2 {
   public static void main(String[] args) throws IOException, InterruptedException, ExecutionException, TimeoutException 
   {
        String ServerURI = "http://localhost:8080/signalr";
        HubConnection Connection = new HubConnection(ServerURI);
        HubProxy HubProxy = Connection.createHubProxy("MyHub");

        HubProxy.on("AddMessage", new SubscriptionHandler2<String, String>() {
          @Override
          public void run(String e1, String e2) {
             System.out.println(e1.toString()+  " -> " +e2.toString());   
          }
        }, String.class, String.class);

        SignalRFuture<Void> con  =Connection.start(new ServerSentEventsTransport(Connection.getLogger())); //Or LongPollingTransport

        con.get(); 

        Scanner inputReader = new Scanner(System.in);
        String line = inputReader.nextLine();
        while (!"exit".equals(line)) {
              HubProxy.invoke("send", "Console", line);
              line = inputReader.next();
         }

        inputReader.close();

        Connection.stop();
   }
}