问Java中的持久客户端-服务器套接字连接
EN

Stack Overflow用户

提问于 2015-10-27 14:59:08

回答 1查看 3.5K关注 0票数 2

我试着用Java建立双向通信。为此，我创建了一个简单的客户机-服务器程序(我甚至不确定这是否是正确的方法，因为它们都被认为是客户端！)。我的要求是，两个程序一旦连接，就应该维护连接，并且能够偶尔发送消息。但是，目前我的套接字一旦发送数据(客户端的main方法已经执行)就终止了。

如何维护连接，以便使其持久？

有什么方法可以使它以一种更适合我的需求的异步方式工作(我不希望服务器总是在while中循环)？

客户端：

package test;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.net.Socket;
import java.net.UnknownHostException;

public class EchoClient2 {
    String serverHostname = new String("127.0.0.1");
    BufferedReader stdIn;
    Socket echoSocket = null;
    PrintWriter out = null;
    BufferedReader in = null;

    public void open(){
        System.out.println("Attemping to connect to host " + serverHostname
                + " on port 9999.");
        try {
            echoSocket = new Socket(serverHostname, 9999);
            out = new PrintWriter(echoSocket.getOutputStream(), true);
            in = new BufferedReader(new InputStreamReader(
                    echoSocket.getInputStream()));
        } catch (UnknownHostException e) {
            System.err.println("Don't know about host: " + serverHostname);
        } catch (IOException e) {
            System.err.println("Couldn't get I/O for " + "the connection to: "
                    + serverHostname);
        }
    }

    public void send(String s){
        out.println(s);
    }

    public String receive(){
        String result = "";
        try {
            result = in.readLine();
            System.out.println("client received: "+result);
            if(result==null)
                return "0";
        } catch (IOException e) {
        }
        return result;
    }


    public static void main(String[] args) {
        EchoClient2 ec = new EchoClient2();
        ec.open();
        ec.send("1");
        ec.receive();
    }
}

服务器：

package test;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.net.ServerSocket;
import java.net.Socket;

public class EchoServer extends Thread {
    protected Socket clientSocket;

    public static void main(String[] args) throws IOException {
        ServerSocket serverSocket = null;

        try {
            serverSocket = new ServerSocket(9999);
            System.out.println("Connection Socket Created");
            try {
                while (true) {
                    System.out.println("Waiting for Connection");
                    new EchoServer(serverSocket.accept());
                }
            } catch (IOException e) {
                System.err.println("Accept failed.");
                System.exit(1);
            }
        } catch (IOException e) {
            System.err.println("Could not listen on port: 9999.");
            System.exit(1);
        } finally {
            try {
                serverSocket.close();
            } catch (IOException e) {
                System.err.println("Could not close port: 9999.");
                System.exit(1);
            }
        }
    }

    private EchoServer(Socket clientSoc) {
        clientSocket = clientSoc;
        start();
    }

    public void run() {
        System.out.println("New Communication Thread Started");

        try {
            PrintWriter out = new PrintWriter(clientSocket.getOutputStream(),
                    true);
            BufferedReader in = new BufferedReader(new InputStreamReader(
                    clientSocket.getInputStream()));

            String inputLine;

            while ((inputLine = in.readLine()) != null) {
                System.out.println("Server received: " + inputLine);
                out.println(inputLine);

                if (inputLine.equals("Bye."))
                    break;
            }

//          out.close();
//          in.close();
//          clientSocket.close();
        } catch (IOException e) {
            System.err.println("Problem with Communication Server");
            System.exit(1);
        }
    }
}

java

sockets

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回答 1

Stack Overflow用户

回答已采纳

发布于 2017-09-21 04:33:31

首先，对迟来的答复表示歉意。我也被困在这个java套接字编程中。我还想打开一个套接字，并保持它为客户端和服务器之间的通信打开。在搜索和阅读了许多书籍和文章之后，我找到了最简单的解决方案。您可以将android生命周期用于

在onStart方法上打开套接字

  protected void onStart() {
        super.onStart();
        new Thread(new Runnable() {
            @Override
            public void run() {
                try {
                    client = new Socket(ip,Integer.valueOf(port) );
                    client.setKeepAlive(true);

                   dataInputStream = new DataInputStream(client.getInputStream());
                  //  readdata=readdata+" "+dataInputStream.readUTF();

                    readdata=dataInputStream.readUTF();
                }
                catch (Exception e){
                    e.printStackTrace();
                }
            }
        }).start();
    }

并关闭onStop方法上的套接字

 protected void onStop() {
        super.onStop();
        try {
            client.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }