客户端-服务器示例不起作用

Question

I studied the Client-Server chapter of a Java book and I copied the code examples for a easy Client-Server interaction.

The server:

package knowledge;

import java.io.*;
import java.net.*;

public class DateTimeServer {

public static void main(String[] args) {
    try {
        int port = Integer.parseInt(args[0]);
        ServerSocket server = new ServerSocket(port);
        System.out.println("DateTimeServer laeuft");
        Socket s = server.accept();
        new DateTimeProtokoll(s).transact();
    } catch (ArrayIndexOutOfBoundsException ae) {
        System.out.println("Aufruf: java DateTimeServer <Port-Nr>");
    } catch (IOException e) {
        e.printStackTrace();
    }

}

}

Protocol:

package knowledge;

import java.io.*;
import java.net.*;
import java.util.*;
import java.text.*;

public class DateTimeProtokoll {
static SimpleDateFormat time = new SimpleDateFormat(
        "´Es ist gerade´H´.´mm´ Uhr.´");
static SimpleDateFormat date = new SimpleDateFormat(
        "´Heute ist´EEEE´, der ´dd.MM.yy´");

Socket s;
BufferedReader vomClient;
PrintWriter zumClient;

public DateTimeProtokoll(Socket s) {
    try {
        this.s = s;
        vomClient = new BufferedReader(new InputStreamReader(
                s.getInputStream()));
        zumClient = new PrintWriter(s.getOutputStream(), true);
    } catch (IOException e) {
        System.out.println("IO-Error");
        e.printStackTrace();
    }
}

public void transact() {
    System.out.println("Protokoll gestartet");
    try {
        zumClient.println("Geben Sie DATE oder TIME ein");
        String wunsch = vomClient.readLine();
        Date jetzt = new Date();

        if (wunsch.equalsIgnoreCase("date"))
            zumClient.print(date.format(jetzt));
        else if (wunsch.equalsIgnoreCase("time"))
            zumClient.println(time.format(jetzt));
        else
            zumClient.println(wunsch + "ist als Kommando unzulaessig!");
        s.close();
    } catch (IOException e) {
        System.out.println("IO-Error");
    }

    System.out.println("Protokoll beendet");
}
}

The Client:

package knowledge;
import java.net.*;
import java.io.*;


public class DateTimeClient {

public static void main(String[] args) {
    String hostName="";
    int port;
    Socket c=null;

    try{
        hostName=args[0];
        port= Integer.parseInt(args[1]);
        c=new Socket(hostName,port);

        BufferedReader vomServer=new BufferedReader(
                                        new InputStreamReader(c.getInputStream()));

        PrintWriter zumServer=new PrintWriter(c.getOutputStream(),true);

        BufferedReader vonTastatur=new BufferedReader(
                                        new InputStreamReader(System.in));
        System.out.println("Server "+ hostName+":"+port+ "sagt:");
        String text=vomServer.readLine();
        System.out.println(text);
        text=vonTastatur.readLine();
        zumServer.println(text);
        text=vomServer.readLine();
        System.out.println(text);

        c.close();
    }
    catch(ArrayIndexOutOfBoundsException ae){
        System.out.println("Aufruf:");
        System.out.println("java DateTimeClient <HostName><PortNr>");
    }
    catch(UnknownHostException ue){
        System.out.println("Kein DNS-Eintrag fuer: "+hostName);
    }
    catch(IOException e){
        System.out.println("IO-Error");
    }



}

}

Here are some notes of my approach and my beliefs. Please disagree on wrong statements stated below:

1)I believe it is no problem to run Client as well as Server on the same (my) computer.

2)I use Eclipse, so I run Eclipse two times in two different workspaces.

3)My input for server program is (run configuration->arguments): 2222

4)My input for the client program is: 2223 my_ip (my_ip is for example 127.0.0.1 I choosed to write my_ip instead because I am not sure if it is dangerous to reaveal my ip in public)

4b) also: "2223" "my_ip"

4c) and: {"2223","my_ip"}

5) also 2222 my_ip (although the figure in my book suggests that the port numbers of client and server should be different, but you never know)

Also I get this very often this error message: Address already in use sounds like client and server port numbers should be different. But I dont know and thats why I ask. Thank you for your help (The page did not like my code so I took screenshots :/)

I replaced the images with code. Sorry for the inconvenience. Restarting first the server with input 2222 and then the client with input 127.0.0.1 2222. After a while the client posts "IO-Error". Its the exception from the clien class (not from protocol class) right? Why is the exception triggered? Is the code working for someone? Thank you

Answer 1

1) Correct. Client and server can be on the same computer, no problem there.

2) Not required, but ok.

3) This will make the server listen on port 2222

4) 127.0.0.1 is just another way of saying "this computer". It is the same as "localhost". Your actual ip is irrelevant to the question anyway, 127.0.0.1 will suffice.

5) Your server is asking on which port to listen (the SOURCE port), your client is asking to which port to connect to (the TARGET port of the client). Of course that should be the SAME port, otherwise the client will try to send a message to port X while the server will listen on port Y.

Imagine the ip as a house address, for example "Mainstreet 12, MyCity". The port would be the appartment number then. Your server occupies appartment 2222, so of course the client needs to try to connect to appartment 2222, otherwise it will not reach any server.

The error is most likely just because you don't actually stop your old server program. Stop it (big red button in eclipse), otherwise it will "occupy" the given port (which will prevent any other program from listing at that port, thus you cannot have two servers running which are both listening on the same port). If we reuse my crude analogy: An appartment cannot contain two servers at the same time, so if one is already in appartment 2222, a second one trying to live there will fail.

Answer 2

I don't know exactly why, but usually server binds host as 127.0.0.1 or localhost or your IP like 192.168.1.100 but if one of that are not listed, than it might be failed to call. See more with netstat.exe on Windows.

The "Address already in use" exception is only because the last Java session still running, could not terminate for some reason, and if you use IDE like Eclipse, it will often happen. :) Make sure that all java thread is terminated (with task manager in Windows). Note that Eclipse is a Java thread too! You would better run the server in console mode..

In real world, native sockets are rarely used, because there some higher level protocol and technology like HTTP, SOAP, WebService, microService, SOA etc. If native socket is not necessary, for example you communicate with microcontroller, you should use these technologies, because these are more robust and others can easily communicate with your interface.