Hi,

I have a mouseover popup using jquery and CSS.

I have the following code in my head:
Code:
<script>
this.screenshotPreview = function(){	
	/* CONFIG */
		
		xOffset = 10;
		yOffset = 30;
		
		// these 2 variable determine popup's distance from the cursor
		// you might want to adjust to get the right result
		
	/* END CONFIG */
	$("a.screenshot").hover(function(e){
		this.t = this.title;
		this.title = "";	
		var c = (this.t != "") ? "<br/>" + this.t : "";
		$("body").append("<p id='screenshot'><img src='"+ this.rel +"' alt='url preview' />"+ c +"</p>");								 
		$("#screenshot")
			.css("top",(e.pageY - xOffset) + "px")
			.css("left",(e.pageX + yOffset) + "px")
			.fadeIn("fast");						
    },
	function(){
		this.title = this.t;	
		$("#screenshot").remove();
    });	
	$("a.screenshot").mousemove(function(e){
		$("#screenshot")
			.css("top",(e.pageY - xOffset) + "px")
			.css("left",(e.pageX + yOffset) + "px");
	});			
};
  $(document).ready(function(){
 		
		screenshotPreview();
});
And call it like so:
Code:
echo "<a href=\"\" class=\"screenshot\" rel=\"images/comingsoon.png\" title=\"$address\"></a>";
I was wondering if it could be more "intelligent"? The problem I have is that is can be out of view and therefore useless. Is there a way to make it always appear in screen space?

Cheers,
Rhys