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Javascript URL参数获取与调用后台API

2012-04-12 10:46 483 查看

第一个页面：index.html

<html>
<body>
<h2>Test</h2>
<hr />
<a href = "/lectureEx.html?rank=00&provider=open163" target = "_blank">Page00</a>
<a href = "/lectureEx.html?rank=01&provider=open163" target = "_blank">Page01</a>
</body>
</html>

第二个页面：lectureEx.html

<html>
<head>
<script>
function showItem()
{
var url = window.location.href;
var paramStr = url.substring(url.indexOf("?") + 1);
var params = paramStr.split("&");
var rank, provider;
for (i = 0; i < params.length; i++)
{
var p = params[i].split("=");
if (p[0] == "rank")
{
rank = p[1];
}
else if (p[0] == "provider")
{
provider = p[1];
}
}

var xmlhttp;
if (window.XMLHttpRequest)
{
xmlhttp = new XMLHttpRequest();
}
else
{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}

xmlhttp.onreadystatechange = function()
{
if (xmlhttp.readyState == 4 && xmlhttp.status == 200)
{
var xmlDoc = xmlhttp.responseXML;
var title = xmlDoc.getElementsByTagName("Title_Original")[0].childNodes[0].nodeValue;
document.getElementById("title").innerHTML = title;

var movies = xmlDoc.evaluate("//Movie", xmlDoc, null, XPathResult.ORDERED_NODE_ITERATOR_TYPE, null);
var m = movies.iterateNext();
var links = "";
while (m)
{
var subtitle = m.getElementsByTagName("fileName")[0].childNodes[0].nodeValue;
var link = m.getElementsByTagName("locationUri")[0].childNodes[0].nodeValue;
links = links + "<a href=\"" + link + "\">" + subtitle + "</a>" + "<br />";
m = movies.iterateNext();
}
document.getElementById("links").innerHTML = links;

var thumbxml = xmlDoc.evaluate("//Thumbnail/locationUri", xmlDoc, null, XPathResult.STRING_TYPE, null);
document.getElementById("thumbnail").setAttribute("src", thumbxml.stringValue);

var summary = xmlDoc.getElementsByTagName("Summary_Short")[0].childNodes[0].nodeValue;
document.getElementById("desc").innerHTML = summary;
}
}

var urltemp = "/crawlerServer/topvideo/single?rank={rank}&datatype=xml&provider={provider}";
var url = urltemp.replace("{rank}", rank).replace("{provider}", provider);
xmlhttp.open("GET", url, true);
xmlhttp.send();
}
</script>
</head>
<body onload = "showItem()">
<div id = "show">
<h2 align = "center" id = "title"></h2>
<img src = "" id = "thumbnail"></img>
<p id = "desc"></p>
<p id = "links"></p>
</div>
</body>
</html>

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