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Thread: pop up menu position
Apr 13, 2004, 14:03 #1
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- Apr 2004
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pop up menu position
<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.
'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?
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?