Java套接字：如何仅在DataInputStream中显示消息一次

Question

I have written a class which already allows me to connect 2 devices using sockets, I also implemented the ability to let each other send a message. I get the received message with a DataInputStream which constantly checks for a message in a while(true){} loop. I log the messages, but they spam the whole console. So I added a variable called lastMsg which checks whether the current message is the last received message, if it is the case, it will not print the message, so the message will only be logged once. The problem is that when a user sends two identical messages, only the first one will obviously be logged.

So is there an other way to check if a message is either send two times, or the message is just the same one sent some seconds ago?

My code to check the received message:

 public void run() {
            Socket other = new Socket();
            try {
            other.connect(new InetSocketAddress("xxxxxx", xxxx), 1000);
            log("Connected to host");
            while (true) {

                DataInputStream in = new DataInputStream(other.getInputStream());
                String msg = in.readUTF();
                if(!lastClientMsg.equals(msg)){
                    log("[HOST-MSG] "+msg);
                    lastClientMsg = msg;
                }


            }
                } catch (IOException e) {
                log("Error at creating client: "+e.toString());
            }

        }

Answer 1

Try using a collection object to store last 20 messages. ArrayList will do fine. check last 20 messages and remove the first one in every new message and add the new one to the collection.

Good Luck.

Answer 2

Using the DataInputStream.readUTF() method should move the message from the input buffer to the string variable, but the buffer is not being cleared for some reason. A few tips:

• Try moving your DataInputStream declaration outside of the while loop. Declaring it every time you check for a message could be your problem.
• Make sure that your code sending the message only sends it once. If by any chance it is being sent in a loop it'll clog up your input stream.

Hope this helps!