Thank you for your help!

On 4月18日, 午後8:29, Jeremy Geerdes <[email protected]> wrote:
> Check out the Feed class' .setNumEntries method. Call 
> feed.setNumEntries(number) before you call feed.load, and you should be good 
> to go.
>
> http://code.google.com/apis/ajaxfeeds/documentation/reference.html#Feed
>
> Jeremy R. Geerdes
> Effective website design & development
> Des Moines, IA
>
> For more information or a project quote:http://jgeerdes.home.mchsi.com
> [email protected]
>
> If you're in the Des Moines, IA, area, check out Debra Heights Wesleyan 
> Church!
>
> On Apr 18, 2010, at 4:49 AM, soujiro0725 wrote:
>
>
>
>
>
> > hi I have a question about Google Ajax Feed API, specifically about
> > "results in XML"
>
> > Google Code playground provides a html + javascript code.  I tried to
> > change the code to get a feed I designate but I only get FOUR results
> > (four titles for this particular code given below).
>
> > Can I possibly get more results?
>
> > For example, let's say, I wish to obtain a list of photos from Picasa
> > Since each folder can contain more than four photos, so the results
> > must be maximal rather than a part of them (unless there is a better
> > way).
> > The code is below but is the exactly same as the one from "Play
> > ground" (http://code.google.com/apis/ajax/playground/?
> > exp=feeds#results_in_xml)
> > ------------------------------------------------------
> > <!--
> >  copyright (c) 2009 Google inc.
>
> >  You are free to copy and use this sample.
> >  License can be found 
> > here:http://code.google.com/apis/ajaxsearch/faq/#license
> > -->
>
> > <!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="http://www.google.com/jsapi?
> > key=ABQIAAAA1XbMiDxx_BTCY2_FkPh06RRaGTYH6UMl8mADNa0YKuWNNa8VNxQEerTAUcfkyrr 
> > 6OwBovxn7TDAH5Q"></
> > 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('item');
> >        // 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
> >        }
> >      }
> >    }
>
> >    function OnLoad() {
> >      // Create a feed instance that will grab Digg's feed.
> >      var feed = new google.feeds.Feed("http://www.digg.com/rss/
> > index.xml");
>
> >      // 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>
> >
> > ---------------------------------------------------------------------------
>
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