Get code to reject username if it exists in the ArrayList.

Question

In a Chatroom code, I have been trying to figure out how to add a code so I can reject client with a username that already exists. I have created an ArrayList where I can store the usernames of the connected clients. But every time I run a client, it accepts a name that I have previously used. This is my code.

if(obj == login) {
  // Need to establish a connection request
  String username = tf.getText().trim();
  clientList.add(username);
  // ignore empty username

  if(username.length() == 0 || username.equals("Your name")){
    JOptionPane.showMessageDialog(null,"Please enter a valid username","Alert Message",JOptionPane.WARNING_MESSAGE);
    return;
  }

  if(clientList.size() > 1 && clientList.contains(username)){
    JOptionPane.showMessageDialog(null,"This username is in use","Alert Message",JOptionPane.WARNING_MESSAGE);
    return;
  }

  // try creating a new Client with GUI
  client = new Client(username, this);
  // test if we can start the Client

  if(!client.start()) 
    return;

  //clientList.add(client);
  tf.setText("");
  tf.setBackground(Color.YELLOW);
  label.setText("Enter your message in the yellow box");
  connected = true;
  login.setEnabled(false);
  logout.setEnabled(true);
  tf.addActionListener(this);
  // Action listener for when the user enter a message
}

Any help please?

Answer 1

As the comment of @Elliott-Frisch, You need to add username to clientList after you check its validity. You may also simplify clientList.size() > 1 && clientList.contains(username)) as clientList.contains(username)

You can just put clientList.add(username); before the line you create the client object:

clientList.add(username);
//try creating a new Client with GUI
client = new Client(username, this);
//test if we can start the Client

Answer 2

if (!clientList.contains(username)){

  clientList.add(username);

}