Question

public static void record(Message  message)//Message is a class 
    {
    try
      {
        BufferedOutputStream buf=new BufferedOutputStream(new      FileOutputStream("E:/kruthika/proj/a.bin"));
        byte[] b =serializer.serialize(message);        
        buf.write(b);
        buf.flush();


      }
    catch(Exception e){System.out.print(e);}
  }

this is a small code which serializes my message object(this object contains some text message) and writes it to a binary file. This works perfectly if the object carries only one word for a text but if many words are sent, obviously it is overwritten in the binary file. So how do i alter the snippet to create multiple files dynamically and then store different message in different file?

Answer 1

you can either pass the fileName-parameter to the function or use a static counter or the current datetime as filename:
shown is the solution with a counter

private static int count = 0;

public static void record(Message  message)//Message is a class 
    {
    try
      {
        BufferedOutputStream buf=new BufferedOutputStream(
          new FileOutputStream("E:/kruthika/proj/" + count + ".bin")
        );
        byte[] b =serializer.serialize(message);        
        buf.write(b);
        buf.flush();

        count++;

      }
    catch(Exception e){System.out.print(e);}
  }