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Get relative and absolute position of the dropped object in JavaScript

Description

The following code shows how to get relative and absolute position of the dropped object.

Example


<!--from w  w w  . ja v  a2 s  .c  o m-->

<html lang="en">
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.10.4/jquery-ui.min.js"></script>
<link type="text/css" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.10.4/themes/smoothness/jquery-ui.css" rel="stylesheet" />
<script type="text/javascript">
$(function() {
var dragOpts = {
helper: "clone",
stop: getNewPos
};

function getNewPos(e, ui) {

e.stopPropagation();
alert(ui.position.top + "px from the top, " + ui.position.left + "px to the left of the original object.");
alert(ui.absolutePosition.top + "px from the top, and " + ui.absolutePosition.left + "px to the left relative to the page.");
}


$("#drag").draggable(dragOpts);
});
</script>
</head>
<body>
<div id="container">
<div id="drag">java2s.com</div>
</div>

</body>
</html>

Click to view the demo

The code above generates the following result.

Get relative and absolute position of the dropped object in JavaScript
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