Determine how to achieve the end return to the previous one else if statement ? ? ?

String username = JOptionPane.showInputDialog (" Please enter your nickname : " ) ;
if (username! = null && username.length ()> 0)
{
System.out.println (username);
JOptionPane.showMessageDialog (a, " enter the nickname :" + username) ;
}
else
{
JOptionPane.showMessageDialog (null, " Please re- enter your nickname ! " ) ;
return;
}
When the input is empty , the system pop-up dialog "Please re-enter your nickname ! ", then how to return "Please enter your nickname : " dialog and continue program execution it ? ? ?
------ Solution ---------------------------------------- ----

  public class DialogTest {
  
  	/**
  	 * @param args
  	 */
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  
  		String username = JOptionPane.showInputDialog("请输入您的昵称:");
  		if (username != null && username.length() > 0) {
  			System.out.println(username);
  			JOptionPane.showMessageDialog(null, "输入的昵称为：" + username);
  		} else {
  			JOptionPane.showMessageDialog(null, "请重新输入您的昵称!");
  			while(username == null || username.length() <=0){
  				username = JOptionPane.showInputDialog("请输入您的昵称:");
  			}
  			System.out.println(username);
  		}
  	}
  
  }
  
  

------ Solution ------------------------------ --------------
while (username == null | | username.length () <= 0) {
username = JOptionPane.showInputDialog (" Please enter your nickname: " ) ;
}

This certainly been waiting for user input ah.

landlord to give you ideas, you can add a tag , such as user input three times as empty, direct end , you think
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is a little complicated
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Thank you ! ! !
But there is a small problem , if the user input is empty again , then the program will not continue, but has been pop-up " Please enter your nickname : ", a deadlock . This is how to do ~ ~ ~
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Thank you ! ! ! Very helpful ! ! !