急,急:这个警告是什么意思--The serializable class chatClient does not declare a static final serialVersionUID field of
我是用一个eclipse来编一个client/server程序,为何在Client程序中出现上述这个警告
我的Client 程序是
import java.io.*;
import java.net.*;
import java.awt.*;
import java.awt.event.*;
public class chatClient extends Frame {
Socket s = null;
DataOutputStream dos = null;
TextField texfid = new TextField();
TextArea taContent = new TextArea();
public void launchFrame() {
setLocation(300, 300);
this.setSize(400, 400);
add(texfid, BorderLayout.SOUTH);
add(taContent, BorderLayout.NORTH);
pack();
this.addWindowListener(new WindowAdapter() {
public void windowClosing(WindowEvent e) {
System.exit(0);
disconnect();
}
});
texfid.addActionListener(new TFListener());
setVisible(true);
connect();
}
public void connect() {
try {
s = new Socket( "127.0.1.1 ", 8888);
System.out.println( "connected ");
} catch (UnknownHostException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
public void disconnect() {
try {
dos.close();
} catch (IOException e) {
e.printStackTrace();
}
}
private class TFListener implements ActionListener {
public void actionPerformed(ActionEvent e) {
String str = texfid.getText().trim();
taContent.setText(str);
texfid.setText( " ");
try {
dos = new DataOutputStream(s.getOutputStream());
dos.writeUTF(str);
dos.flush();
} catch (IOException e1) {
e1.printStackTrace();
}
}
}
public static void main(String[] args) {
new chatClient().launchFrame();
}
}
我的Server程序是:
import java.net.*;
import java.io.*;
class chatServer2 {
boolean bConnected=false;
DataInputStream dis=null;
public static void main(String[] args) {
new chatServer2().start();
}
public void start(){
ServerSocket ss=null;
try {
ss=new ServerSocket(8888);
bConnected=true;
}catch(BindException eb){
System.out.println( "port in use..... ");
System.out.println( "please close other program and continue ");
}catch(IOException ei){
ei.printStackTrace();
}
try {
Socket s = ss.accept();
client c=new client(s);
new Thread(c).start();
} catch (IOException e) {
e.printStackTrace();
}finally{
try {
ss.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
class client implements Runnable{
private Socket s= null;
private DataInputStream dis=null;
boolean bConnected=false;
public client(Socket s){
this.s=s;
try {
dis=new DataInputStream(s.getInputStream());
} catch (IOException e) {
e.printStackTrace();
}
bConnected=true;
}
public void run() {
try{
while(bConnected){
dis=new DataInputStream(s.getInputStream());
String str=dis.readUTF();
System.out.println(str);
}
}catch(SocketException ee){
System.out.println( "the client is over! ");
}catch (IOException e){
e.printStackTrace();
}finally{
try {
if(dis !=null) dis.close();
if(s !=null) s.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
}
[解决办法]
这个好像是跟对象序列化有关，一般情况下不需要理会的
[解决办法]
我遇到，也是不予理会。
