Question

我想从此链接中提取XML

http://212.12.182.204:8090/NewsService.svc/LatestNews

使用JavaScipt，我已经完成并搜索了大量代码，但我认为错误存在于链接或提取本身，我不确定，因为我是新手。

谢谢

编辑部分：我在谷歌代码游乐场使用的代码：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
  <head>
    <meta http-equiv="content-type" content="text/html; charset=utf-8"/>
    <title>Google AJAX Search API Sample</title>
    <script src="//www.google.com/jsapi?key=AIzaSyA5m1Nc8ws2BbmPRwKu5gFradvD_hgq6G0" type="text/javascript"></script>
    <script type="text/javascript">
/*
*  How to receive results in XML.
*/

google.load("feeds", "1");

// Our callback function, for when a feed is loaded.
function feedLoaded(result) {
  if (!result.error) {
    // Get and clear our content div.
    var content = document.getElementById('content');
    content.innerHTML = '';

    // Get all items returned.
    var items = result.xmlDocument.getElementsByTagName('News');
    // Loop through our items
    for (var i = 0; i < items.length; i++) {
      var item = items[i];

      // Get the title from the element.  firstChild is the text node containing
      // the title, and nodeValue returns the value of it.
      var title = item.getElementsByTagName('Title')[0].firstChild.nodeValue;

      content.appendChild(document.createTextNode(title)); // Append the title to the page
      content.appendChild(document.createElement('br')); // Add a new line
    }
  }
  else
  {
    alert(result.error.code);
  }
}

function OnLoad() {
  // Create a feed instance that will grab Digg's feed.
  var feed = new google.feeds.Feed("http://212.12.182.204:8090/NewsService.svc/LatestNews");

  // Request the results in XML
  feed.setResultFormat(google.feeds.Feed.XML_FORMAT);

  // Calling load sends the request off.  It requires a callback function.
  feed.load(feedLoaded);
}

google.setOnLoadCallback(OnLoad);
</script>
  </head>
  <body style="font-family: Arial;border: 0 none;">
    <div id="content">Loading...</div>
  </body>
</html>

如何从JavaScript中的此链接中提取XML？

0 个答案: