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slide php code

Posted on 2014-04-10
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Last Modified: 2014-04-10
I got the code below from http://unslider.com/ and it does not seem to work. May be I did something wrong. I am very beginner in programming php so any help will be greatly appreciated.

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>

<style type="text/css">
<!--
.banner { position: relative; overflow: auto; }
    .banner li { list-style: none; }
        .banner ul li { float: left; }
-->
</style>

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script src="//ajax.googleapis.com/ajax/libs/jqueryui/1.10.4/jquery-ui.min.js"></script>

<script src="//unslider.com/unslider.min.js"></script>

<script type="text/javascript">
		$(function() {
			$('.banner').unslider();
		});
</script>

</head>

<body>
<div class="banner">
    <ul>
        <li><img src="images/banner1.jpg" width="960" height="202" /></li>
		<li><img src="images/banner_about.jpg" width="960" height="202" /></li>

    </ul>
</div>
</body>
</html>

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Question by:leblanc
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käµfm³d   👽 earned 500 total points
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Where is the "http:" in lines 15, 16, and 18?

i.e.

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.10.4/jquery-ui.min.js"></script>

<script src="http://unslider.com/unslider.min.js"></script>

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by:leblanc
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ahhh... Thank you
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