<TABLE onMouseOver=SubMenu('S1');Swap(this) onMouseOut=Swap(this) class='Menu' id='T1'
style='Z-INDEX: 1; LEFT: 10px; POSITION: absolute; TOP: 50px' cellSpacing='0' cellPadding='3' width='150'height='20' border='0'>


<TABLE class='SubMenu' id='S1' style='visibility: hidden; Z-INDEX: 2; LEFT: 161px; POSITION: absolute; TOP: 50px' cellSpacing='0' cellPadding='3'width='150' border='0'>

What do I have to do so that menu and the sub menu are not fixed but is relative to the screen resolution and browser width and height.

Two options:

'To position based on screen width, height and res use:
document.body.clientHeight or document.body.clientWidth to get the size of the browser window. Or use window.screen.availWidth & window.screen.availHeight to get the available screen height and width.

To get app res use window.screen.width & window.screen.height.'

what code can I use to implement this?

Another option:

I can remove the left, top and position absolute for my parent menu so that it is always in the right place along with the rest of my site content. but I cannot do that for my child menu which is a pop up menu. I need the child menu to be positioned based on the xpos and ypos of my parent menu. How can I do that?