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Editorial Team
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Asked: 2026-06-04T05:39:33+00:00

I am using a script to load news from different sources, using Google AJAX

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I am using a script to load news from different sources, using Google AJAX feed API. How can I get the description of an entry? Below is an hello world program:

<html>
  <head>
    <script type="text/javascript" src="https://www.google.com/jsapi"></script>
    <script type="text/javascript">

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

    function initialize() {
      var feed = new google.feeds.Feed("http://news.google.com/?output=rss");
      feed.load(function(result) {
        if (!result.error) {
          var container = document.getElementById("feed");
          for (var i = 0; i < result.feed.entries.length; i++) {
            var entry = result.feed.entries[i];
            var div = document.createElement("div");
            div.appendChild(document.createTextNode(entry.title));
            container.appendChild(div);
          }
        }
      });
    }
    google.setOnLoadCallback(initialize);

    </script>
  </head>
  <body>
    <div id="feed"></div>
  </body>
</html>

How can I get the description using the entry object??? I am using the google URL – http://news.google.com/?output=rss for RSS feeds in XML format. I want the “Description” part. How can I get that

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1 Answer

  1. Editorial Team

    You can get the description, but you can’t use the JSON format and the entry object to do it. If you read the feed parameters at https://developers.google.com/feed/v1/devguide carefully, you’ll see that description is not a field it returns at the entry level – just at the feed level.

    To do it, you need to request the feed in XML format, and then load the individual nodes, including description. Here’s the relevant snippet I’ve used to do it – change the formatting etc. as you need.

    function initialize() {
   var feed = new google.feeds.Feed("http://myblog.com/blog/feed/");
   feed.setResultFormat(google.feeds.Feed.XML_FORMAT);
   feed.load(function(result) {
   if (!result.error) {
    var items = result.xmlDocument.getElementsByTagName('item');
    item = items[0];

    //build each element
    var title = document.createElement("h4");
    title.innerHTML = item.getElementsByTagName('title')[0].firstChild.nodeValue;

    var content = document.createElement("p");
    content.innerHTML = item.getElementsByTagName('description')[0].firstChild.nodeValue;

    href = item.getElementsByTagName('link')[0].firstChild.nodeValue;
   }
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