我的目标是点击下拉菜单(div id = cssmenu)元素并在使用AJAX中显示数据,所以我制作了一个小的js代码,但我不知道如何让它加载到div我提到并删除了div之前显示的内容。
HTML
<body>
<div class="container">
<div class="row">
<div class="col-md-3" >
<div id='cssmenu'>
<ul>
<li class='active'><a href='#' data-url="http://localhost/bioinformatica/Main_page/Quick_search.html#qhelp"><span>Home</span></a></li>
<li class='has-sub'><a href='#'><span>About</span></a>
<ul>
<li><a href='#'><span>Project</span></a></li>
<li class='last'><a href='#'><span>Team</span></a></li>
</ul>
</li>
<li class='last'><a href='#'><span>News</span></a></li>
</ul>
</div>
</div>
<div class="col-md-9">
<div id="tabs">
<ul class="nav nav-tabs" id="prodTabs">
<li class="active"><a class="clickableLink" href="#tab_quick" data-url="http://localhost/bioinformatica/Main_page/Quick_search.html#qhelp">Quick Search</a></li>
<li><a class="clickableLink" href="#tab_advanc" data-url="#">Advanced Search</a></li>
<li><a class="clickableLink" href="#tab_struct" data-url="something3.txt">Structure Search</a></li>
</ul>
<div class="tab-content">
<div id="tab_quick" class="tab-pane active"></div>
<div id="tab_advanc" class="tab-pane active"></div>
<div id="tab_struct" class="tab-pane active"></div>
</div>
</div>
</div>
</div>
<div class="clearfix visible-lg"></div>
</div>
</div>
</body>
</html>
的Javascript
$( document ).ready(function() {
$('#cssmenu a').click(function (e) {
e.preventDefault();
var url = $(this).attr("data-url");
var href = this.hash;
var pane = $(this);
// ajax load from data-url
$(href).load(url,function(result){
pane.tab('show');
});
});
});
例如在菜单中,当我点击Home时,我希望它在div id = tabs中加载它的data-url attr,即用Home data-url替换所有内容。谢谢 !
答案 0 :(得分:1)
您可以尝试以下操作:
$( document ).ready(function() {
$('#cssmenu a').click(function (e) {
e.preventDefault();
var url = $(this).data('url'); // this is the correct syntax
var pane = $(this);
$('#tabs').load(url, function(result){ // load the content directly to #tabs
pane.tab('show'); // display the tab
});
});
});