How do you pass a feed URL into the widget?

thanks,

ROb

To pass a feed URL to a widget that is set to use dynamic content, simply add the following parameter to the code snippet after the API token:

&xmlUrl=[XMLURL]

As an example, here is a simple widget that makes a call to dynamic content. The parameter only needs to be added to the snippet code that is included on your page where the widget will reside; you do not have to make any changes through Editor's Desk:

<script src="http://nmp.newsgator.com/NGBuzz/Buzz.ashx?buzzId=56443&apiToken=760F1E2C6BFD4C07866316A15F7A127F&xmlurl=http%3A%2F%2Fnewsrss.bbc.co.uk%2Frss%2Fnewsonline_world_edition%2Ffront_page%2Frss.xml" type="text/javascript" language="javascript"></script>

There are two important things to make note of when utilizing dynamic content:

1) The XML URL must be URI encoded (as you can see above, http:// is seen as http%3A%2F%2F). Here is a free utility for encoding your feed URLs: http://lachy.id.au/dev/mozilla/sidebar/URI/uri-dec-enc.html
Simply input the URL and hit "Encode."

2) The feed that is referred to in the parameter must exist in our system. If you see that no posts are showing up for a particular feed, just add it to a widget (you can remove it later as needed) or subscribe to it through any NewsGator service. Once this is done our system will recognize the feed and keep the content fresh.

Let me know if you have any other questions!