My page is rather simple and direct. I first call in external PHP pages as elements within my page using DIV tags. I then hide the div’s that aren’t needed on the home page and then slide into view the wanted “page” and hide the unwanted.
What I would like to do is slide in the wanted content and slide out what ever content is currently active/visible. I know I can just continue down the line what I’ve already begun doing but the entire document would be SO bloated. Any ideas? Thanks!
<link href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/themes/base/jquery-ui.css" rel="stylesheet" type="text/css"/>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.5/jquery.min.js"></script>
<script src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/jquery-ui.min.js"></script>
<script>
$(document).ready(function(e) {
$(function(){
$("div.design, div.marketing").hide();
});
});
$(document).ready(function() {
$("h3.home_button").click(function () {
$("div.home").show("slide", { direction: "left" }, 500);
$("div.design").hide();
$("div.marketing").hide();
});
$("h3.design_button").click(function () {
$("div.design").show("slide", { direction: "left" }, 500);
$("div.home").hide();
$("div.marketing").hide();
});
$("h3.marketing_button").click(function () {
$("div.marketing").show("slide", { direction: "left" }, 500);
$("div.home").hide();
$("div.design").hide();
});
});
</script>
</head>
<body>
<div class="main">
<div class="header">
<?php include("header.php");?>
</div>
<div class="menu">
<?php include("menu.php");?>
</div>
<div class="content">
<?php include("content_home.php");?>
<?php include("content_design.php");?>
<?php include("content_marketing.php");?>
</div>
<div class="footer"><?php include("footer.php");?></div>
</div>
You could use the
data attributesand access it by thedata()jquery method.in your HTML
and in your javascript
Here is an exemple fiddle
Also you are calling ready method twice