Dear all,
I created a pop up window by using "Ext.window".When i click the button it pops up a window.I want to set the location of popup window by looking at mouse clicked position.For example When i click button,the popup window must be opened near that button.How can i do this?Is there anyway to set popup windows css to "position:fixed"?
Thanks for advance

Code:

if (!win.rendered) {
  win.render(Ext.getBody());
}
win.setPagePosition(e.getXY());
win.show();

It didn't work :-(

What didn't work?

Here is my code:
win = new Ext.Window({
applyTo : 'example',
layout : 'auto',
autoScroll : 'true',
width : 650,
height : 400,
closeAction : 'hide',
plain : true,
html : oXHR.responseText,
title:"example title",
buttons: [{
text : ' Add >> ',
handler : function() {
func1();
win.hide();
}
}]
});
if (!win.rendered) {
win.render(Ext.getBody());
}
win.setPagePosition(e.getXY());
win.show();

this code exist in a functions that belongs a button function.This window always pop up at center of page.I want to set window location manually.

OK, you need to use setPosition instead of setPagePosition, or just assign an x and an y to the window.

when i use x and y tags ,the window sets its position according to the top of the page.For example ,if the page is very long and i use the scroll and go to bottom of the page ,after i press that button scroll goes up and the window set its position at the top (if x:,y:0).

It helps me a lot..Really thank full to you

Not necessarily. I always set "constrain:true" to all my EXT windows so that they never flow outside of their parent, even if the x&y coordinates would normally push it outside...