sac*_*are 15 html javascript jquery
我是Jquery的新手.我正在尝试从"sampleXML.xml"文件中读取数据并在Html"li"元素中显示该数据.到目前为止我所做的是,我创建了如下html文件:文件名 - "Cloudtags.html":
<!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" />
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"> </script>
<script src=Cloudtags.js></script>
<title>Css Globe: tag clouds</title>
<link rel="stylesheet" type="text/css" href="Cloudtags.css">
<script type="text/javascript" src="js/jquery.js"></script>
</head>
<body>
<div id="container">
<script type="text/javascript" src="http://cssglobe.com/ads/blogsponsor.js"></script>
<div id="side">
<div class="tags">
<ul class="cld">
<li class="tag1" id="java"><a href="https://www.google.com">google</a></li>
<li class="tag2"><a href="#">Chiessy</a></li>
<li class="tag3"><a href="#">sitemap</a></li>
<li class="tag2"><a href="#">Sales</a></li>
<li class="tag4" ><a href="#">Gohome</a></li>
<li class="tag1"id="temp"><a href="#">Movies</a></li>
<li class="tag1"><a href="#">It Jobz</a></li>
<li class="tag5"><a href="#">Alza</a></li>
<li class="tag2"><a href="#">Sea food</a></li>
<li class="tag1"><a href="#">Hospital</a></li>
<li class="tag3"><a href="#">Smart phone</a></li>
<li class="tag4"><a href="#">Pizza </a></li>
<li class="tag1"><a href="#">Aerobics</a></li>
<li class="tag5"><a href="#">Yahoo...</a></li>
<li class="tag1"><a href="#">Anti-Virus</a></li>
<li class="tag2"><a href="#">Travel</a></li>
</ul>
</div>
</div>
<div id="xmldata"></div>
</div><br>
</body>
</html>
这是我的.js文件:
$(document).ready(function() {
var nm;
$.ajax({
type: "GET" ,
url: "sampleXML.xml" ,
dataType: "xml" ,
success: function(xml) {
$(xml).find('person').each(function() {
nm= $(this).text()
$("#temp").html(nm);
}
}
});
});
我的xml文件如下:
<?xml version='1.0' ?>
<doc>
<person>
<name>sachin</name>
<age>21</age>
</person>
<person>
<name>Akash</name>
<age>18</age>
</person>
</doc>
但这不起作用.我是否需要为"$ .ajax"链接一些外部文件.所以,请告诉我我在哪里弄错了..提前致谢 ..
Ris*_*Php 28
我想你想这样,DEMO
var xmlDoc = $.parseXML( xml );
var $xml = $(xmlDoc);
var $person = $xml.find("person");
$person.each(function(){
var name = $(this).find('name').text(),
age = $(this).find('age').text();
$("#ProfileList" ).append('<li>' +name+ ' - ' +age+ '</li>');
});
Chr*_*les 13
您只需将XML文件读取为dataType:"xml",它将重新调整已解析的xml对象.你可以将它用作jquery对象并找到任何东西或循环抛出它等等.
$(document).ready(function(){
$.ajax({
type: "GET" ,
url: "sampleXML.xml" ,
dataType: "xml" ,
success: function(xml) {
//var xmlDoc = $.parseXML( xml ); <------------------this line
//if single item
var person = $(xml).find('person').text();
//but if it's multible items then loop
$(xml).find('person').each(function(){
$("#temp").append('<li>' + $(this).text() + '</li>');
});
}
});
});