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Frederic Camara
4,344 PointsWhere do I identify the target attr. var $external = $('<a class="external"></a>') $external.attr('target',"_blank");
I'm not sure what I did wrong here. It is apparently not passing in a value for the target attribute. Am I missing a step or is my syntax improper?
var $external = $('<a class="external"></a>')
$external.attr('target',"_blank");
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" href="css/style.css" type="text/css" media="screen" title="no title" charset="utf-8">
<title>Links Page</title>
</head>
<body>
<h1>Links</h1>
<ul>
<li><a href="http://google.com" class="external">Google</a></li>
<li><a href="http://yahoo.com" class="external">Yahoo</a></li>
</ul>
<script src="//code.jquery.com/jquery-1.11.0.min.js" type="text/javascript" charset="utf-8"></script>
<script src="js/app.js" type="text/javascript" charset="utf-8"></script>
</body>
</html>
1 Answer

Steven Parker
228,077 PointsYou're not creating new elements here, you are selecting existing ones. The JQuery argument will look like a CSS selector, which in this case would be a.external
(for links which have the external class).
SPOILER ALERT
Here's the corrected code:
var $external = $("a.external")
$external.attr('target', "_blank");
This can also be combined into one line:
$("a.external").attr('target', "_blank");