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grace . But that is only useful for the latest launch .
I do not know where the problem is , even if the old press the Send button and read the dialog box is the latest input .
is not your dialog object references wrong
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send your program. Must be your code appears concurrent state inconsistency. A " thread closed" , or a " lock "
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I think the landlord is not a programming problem . After the first connection is established , there will be a Socket object , you should first use the Socket connection at the same time , the user ID sent me , temporarily use A, B logo, then you should save a two -dimensional structure , it is best to HashMap this , numbered do Key, Socket doing Value, so you send a message by parsing the goals, such as , A sent B, sending number is A, the recipient number is B, you parse content, taken from the Map B corresponds in the socket object , and then to B hair. This order fishes . You are not confused into mass news ?
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[/ code] Since the source code of several files I do not fully made , only made the relevant section.
open ports:
while(true)
{
ServerSocket server_socket=new ServerSocket(i1); //i1是端口号
server threadServer=new server(server_socket.accept());
threadServer.start();
}
server class :
import java.io.*;
import java.net.*;
public class server extends Thread
{
int chatcreat;
private Socket myClient;
public server(Socket c)
{
this.myClient=c;
chatcreat=0;
}
public void run()
{
try
{
BufferedReader in=new BufferedReader(new InputStreamReader(myClient.getInputStream()));
PrintWriter out=new PrintWriter(myClient.getOutputStream());
System.out.println("Wait for client to send string...");
//System.out.println("chatcreat="+chatcreat);
String name=in.readLine();
if((!name.equals(""))&&chatcreat==0)
{
chatcreat++;
new chatinter(name,myClient).start(); //在客户端发给服务器的第一条消息前先把客户端自定义的昵称发到服务器上。chatinter是对话界面
}
}
catch(IOException ex){}
finally{}
}
}
interface displays the message part:
try
{
BufferedReader in=new BufferedReader(new InputStreamReader(myClient.getInputStream()));
out=new PrintWriter(myClient.getOutputStream());
while(true)
{
String str=in.readLine();
a1.append(tit+":"+str+"\n");//a1是JTextArea,tit是前面传过来的客户端的昵称
}
}
catch (Exception e1)
{
System.out.println(e1);
}
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still can not figure out why. . Do not sink ah
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which Socket input, processing, then sent out the same Socket ouput
ServerSocket server_socket=new ServerSocket(i1); //i1是端口号
while(true)
{
//放在此处会创建多个ServerSocket
server threadServer=new server(server_socket.accept());
threadServer.start();
}
unreasonable strings
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add newly discovered . After the first two 1st person personal connection interface on the failure , not a message , press the button does not respond , the message will not be updated. I'll give the first person to send a message is also displayed in the interface of the second person .
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add newly discovered . After the first two 1st person personal connection interface on the failure , not a message , press the button does not respond , the message will not be updated. I'll give the first person to send a message is also displayed in the interface of the second person .
The reason is : the code each time before you create a new ServerSocket object , therefore, only the latest Socket effective access
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upstairs positive solution, as long as the first one in while truee while , then moved to the outside like a
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add newly discovered . After the first two 1st person personal connection interface on the failure , not a message , press the button does not respond , the message will not be updated. I'll give the first person to send a message is also displayed in the interface of the second person .
The reason is : the code each time before you create a new ServerSocket object , therefore, only the latest Socket effective access
Then how can I solve it .
do so the first two individual interface will not come out . .
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if the server class definition into the class declaration , the two people connect , the server to the client a message in the text field , then enter the information could not be read , press the " send" button to send the past is the empty message is displayed in the client touches inside an interface . Client 1 , then the server will display a message to the server to the client 2 interface , the content is normal. Client 2 has no effect
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add newly discovered . After the first two 1st person personal connection interface on the failure , not a message , press the button does not respond , the message will not be updated. I'll give the first person to send a message is also displayed in the interface of the second person .
The reason is : the code each time before you create a new ServerSocket object , therefore, only the latest Socket effective access
Then how can I solve it .
do so the first two individual interface will not come out . .
interface and ServerSocket is not related !
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add newly discovered . After the first two 1st person personal connection interface on the failure , not a message , press the button does not respond , the message will not be updated. I'll give the first person to send a message is also displayed in the interface of the second person .
The reason is : the code each time before you create a new ServerSocket object , therefore, only the latest Socket effective access
Then how can I solve it .
do so the first two individual interface will not come out . .
interface and ServerSocket is not related !
interface is connected to receive the first two people will come out ah.
server threadServer = new server (server_socket.accept ()); sentence do not get out while receiving a client can only Mody
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landlord to see me examples Click to download page , I have done this and saved user names and a list of yes yes corresponding connection , so very simple
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landlord to see me examples Click to download page , I have done this and saved user names and a list of yes yes corresponding connection , so very simple
Thank you . But I was trying to figure out my main problem lies in what this program . . I made a paste to another program Quanfa Come on
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add newly discovered . After the first two 1st person personal connection interface on the failure , not a message , press the button does not respond , the message will not be updated. I'll give the first person to send a message is also displayed in the interface of the second person .
The reason is : the code each time before you create a new ServerSocket object , therefore, only the latest Socket effective access
Then how can I solve it .
do so the first two individual interface will not come out . .
interface and ServerSocket is not related !
interface is connected to receive the first two people will come out ah.
server threadServer = new server (server_socket.accept ()); sentence do not get out while receiving a client can only Mody
First of all:
get out the ServerSocket server_socket = new ServerSocket (i1); / / i1 is the port number
rather than accept
Are you sure ?
ServerSocket to listen for client connection requests generate a Socket
no direct correlation between different Scoket
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amount . I've changed like this:
public void run()
{
try
{
i1=Integer.parseInt(t1.getText());
name=t2.getText();
if(name.equals(""))
System.out.println("请输入你的昵称！");
else
{
server_socket=new ServerSocket(i1);
indexset.connectsucceed();
//server threadServer;
while(true)
{
//System.out.println("Wait for client to connect...");
Socket a=server_socket.accept();
//server threadServer=new server(server_socket.accept());
server threadServer=new server(a);
threadServer.start();
}
}
}
catch(Exception e1){ System.out.println("ServerSocket启动失败!");};
}
but still not resolved . . I put all the procedures are uploaded http://download.csdn.net/download/ww01055/6569887
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you are connected to each client initiates a chat box ?
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grace . But that is only useful for the latest launch .
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grace . But that is only useful for the latest launch .
I do not know where the problem is , even if the old press the Send button and read the dialog box is the latest input .
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[quote = 21st floor victor_woo reply quote :
is not your mistake dialog object references
egg pain . Really are. Are you such a reminder found written in the text field of the static type . . . No wonder . Hard upstairs it. Alas
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