我正在尝试对即时消息系统进行编码......最初，我是这样做的，一旦我开始工作，我就会添加GUI。
一旦客户端向服务器发送消息，服务器就应该将其显示给所有其他客户端。我怎样才能做到这一点？我一直在尝试一些东西，但它一直只显示给发送消息的客户端......
提前致谢！
服务器
import java.io.*;
import java.net.*;
class Server {
//one per server
static int port = 3000;
private int backlog = 100;
ServerSocket main;
static DataOutputStream dataOut;
static DataInputStream dataIn;
static String scannerMessage;
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static class MailServer extends Thread {
//one per client
static int index;
String name = Client.name;
public MailServer(int index, DataInputStream in, DataOutputStream out) {
Server.dataIn = in;
Server.dataOut = out;
this.index = index; // thread index, one per client
}
public void run() {
while (true) {
try {
String receivedMessage = dataIn.readUTF();
//print receivedMessage to all clients
} catch (Exception e) {
break;
}
}
}
}
public Server(int port) throws Exception {
this.main = new ServerSocket(port);
}
// start a serve
public void serve() throws Exception {
int index = 1;
while (true) {
Socket socket = this.main.accept();
InputStream in = socket.getInputStream();
OutputStream out = socket.getOutputStream();
DataInputStream dataIn = new DataInputStream(in);
DataOutputStream dataOut = new DataOutputStream(out);
// handle the connection
// keep reading using an infintite loop
System.out.println("Handling connection to Client " + index + "...");
(new MailServer(index, dataIn, dataOut)).start();
index += 1; // add one every time a new client is added
}
}
public static void main(String[] args) throws Exception {
Server s = new Server(port);
System.out.println("Serving....");
s.serve();
}
}
客户端
import java.io.*;
import java.net.*;
class Client {
static String hostname = "127.0.0.1";
static int port = Server.port;
static Socket socket;
static String name;
static class Sender extends Thread {
DataOutputStream dataOut;
public Sender(DataOutputStream dataOut) {
this.dataOut = dataOut;
}
public void run() {
while(true) {
//get a message from the user
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
try {
String message = br.readLine();
dataOut.writeUTF(message);
dataOut.flush();
} catch(Exception e) {
break;
}
}
}
}
static class Receiver extends Thread {
DataInputStream dataIn;
public Receiver(DataInputStream dataIn) {
this.dataIn = dataIn;
}
public void run() {
while(true) {
try {
//RECEIVE A MESAGE FROM THE SERVER (ending in \n)
String msg = dataIn.readUTF();
while (msg != null) {
System.out.println(msg);
msg = dataIn.readUTF();
}
} catch(Exception e) {
break;
}
}
}
}
//client will require host name and the port
public Client(String hostname, int port) throws Exception {
socket = new Socket(hostname, port);
}
public void connect() throws Exception {
InputStream in = socket.getInputStream();
OutputStream out = socket.getOutputStream();
DataInputStream dataIn = new DataInputStream(in);
DataOutputStream dataOut = new DataOutputStream(out);
//handle the connection
System.out.println("Handling connection to server...");
Thread sender = new Sender(dataOut);
Thread receiver = new Receiver(dataIn);
sender.start();
receiver.start();
sender.join();
receiver.join();
System.out.println("Client " + Server.MailServer.index);
System.out.println("----------------------");
}
public static void main(String[] args) throws Exception {
Client c = new Client(hostname, port);
c.connect();
}
}
更新：我创建了一个包含所有MailServer对象的列表，然后通过它们迭代将消息发送给所有客户端，正如JP Moresmau建议的那样......但现在第一个发送内容的客户端收到所有输出。为什么是这样？我该怎么解决它？谢谢，对不起，如果我的问题看起来太明显或愚蠢，我还是一个Java菜鸟：（
服务器 - 更新
package csci2020_assignment51;
import java.io.*;
import java.net.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
class Server {
//one per server
static int port = 3000;
private int backlog = 100;
ServerSocket main;
static DataOutputStream dataOut;
static DataInputStream dataIn;
static String scannerMessage;
static List<MailServer> mailServers = Collections.<MailServer>synchronizedList(new ArrayList<MailServer>());
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static class MailServer extends Thread {
//one per client
static int index;
String name = Client.name;
public MailServer(int index, DataInputStream in, DataOutputStream out) {
Server.dataIn = in;
Server.dataOut = out;
this.index = index; // thread index, one per client
}
public void run() {
while (true) {
try {
String receivedMessage = dataIn.readUTF();
String outputMessage = "Client " + index + " said: " + receivedMessage;
//print receivedMessage to all clients
for (MailServer mailserver : mailServers) {
dataOut.writeUTF(outputMessage);
}
} catch (Exception e) {
break;
}
}
}
}
public Server(int port) throws Exception {
this.main = new ServerSocket(port);
}
// start a serve
public void serve() throws Exception {
int index = 1;
while (true) {
Socket socket = this.main.accept();
InputStream in = socket.getInputStream();
OutputStream out = socket.getOutputStream();
DataInputStream dataIn = new DataInputStream(in);
DataOutputStream dataOut = new DataOutputStream(out);
// handle the connection
// keep reading using an infintite loop
System.out.println("Handling connection to Client " + index + "...");
MailServer mailServer = new MailServer(index, dataIn, dataOut);
mailServer.start();
mailServers.add(mailServer);
dataOut.writeUTF("Client " + index);
index += 1; // add one every time a new client is added
}
}
public static void main(String[] args) throws Exception {
Server s = new Server(port);
System.out.println("Serving....");
s.serve();
}
}
答案 0 :(得分：2)
拥有您创建的所有MailServer对象的静态列表
static List<MailServer> servers=Collections.<MailServer>synchronizedList(new LinkedList<MailServer>);
...
MailServer s=new MailServer(index, dataIn, dataOut);
servers.add(s);
s.start();
然后当其中一个收到消息时，循环遍历它们，并且对于所有人都期望接收者，将消息写入其输出。
代码中的一个大问题是dataOut和dataIn是唯一的！您需要将它们移动到MailServer类。每个MailServer与一个特定的客户端进行通信，因此需要拥有自己的流。
static class MailServer extends Thread {
DataOutputStream dataOut;
DataInputStream dataIn;
您的通知循环变为：
for(MailServer mailServer:mailServers){
if (mailServer!=this){
mailServer.dataOut.writeUtf(...);
}
}
我也不明白你希望如何在服务器中获取Client.name，因为客户端在另一台机器上运行...暂时摆脱它。