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Keeping a java window (JDialog) 'always on top'

Posted on 1998-11-06
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Last Modified: 2012-06-27
Is it possible for a java window to be forced to remain the topmost window?

For my purposes this app will always only run under Win95/98/NT so a native method approach is acceptable, and will work.  However, to do it that way I need a guaranteed way of getting the HWND for the window which I have not yet found.  Calling GetForegroundWindow won't do since there is no way to guarantee that the window is the foreground window when the call is made.
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Question by:mrgentry
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2 Comments
 
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Accepted Solution

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gadio earned 100 total points
ID: 1227034
mrgentry, I have attached here a sample of code that does that. Note however, that on w95/NT, the window will be on op only if the aplication is active. Sothing that has to do with the way that the JVM on windows is implemented. See if you can use it.
Regarding your second question - the best way I found to find the window was to assign a unique name for each window (in the title), and the in the native method go throuh all the windows until you find the one with the expected name. However, I hope this code will do the trick for you. If you need more explenation, ask.



--------------------------------

   import java.awt.*;

   public class alwaysOnTop {

       /**
        * Constructor.
        */
       public alwaysOnTop () {
       }

       public static void main(String args[]) {
           myframe mf1 = new myframe();
           myframe mf2 = new myframe();
           myframe mf3 = new myframe();
           myframe mf4 = new myframe();
           ontopframe tf = new ontopframe();
           mf1.add(new Label("b1"));
           mf1.show();
           mf2.add(new Label("b2"));
           mf2.show();
           mf3.add(new Label("b3"));
           mf3.show();
           mf4.add(new Label("b4"));
           mf4.show();
           tf.add(new Label("Alway on top"));
           tf.show();
       }
   }

   class myframe extends Frame {
        public myframe() {
           setLayout( new FlowLayout());
        }

        public Dimension minumumSize() {
           return new Dimension( 200, 200 );
        }

        public Dimension preferredSize() {
           return new Dimension( 200, 200 );
        }

        public Dimension getPreferredSize() {
           return preferredSize();
        }
   }

   class ontopframe extends Frame implements Runnable {
        public final static int TO_TOP_TIME = 100; // Max time to top in milliseconds.
        Thread top_thread;
        boolean stop = false;

        public ontopframe() {
           setLayout( new FlowLayout());
           top_thread = new Thread(this);
           top_thread.start();
        }

        public Dimension minumumSize() {
           return new Dimension( 200, 200 );
        }

        public Dimension preferredSize() {
           return new Dimension( 200, 200 );
        }

        public Dimension getPreferredSize() {
           return preferredSize();
        }

        public void run() {
           while(!stop) {
               toFront();
               try{
                   top_thread.sleep(TO_TOP_TIME);
               } catch ( Exception e ){}
           }
        }
   }
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by:mrgentry
ID: 1227035
I would really rather not have a thread just sitting there throwing the window to the front constantly and would much prefer to take the 2nd approach of searching for the widow.
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