How to open a URL in widget and avoid the bug of widget.openURL

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Compatibility: Web Runtime in S60 3rd Edition, Feature Pack 2

Bug

How to open a URL? It's recommended to use this one: widget.openURL(url); But it showed error message and couldn't work for the widget.openURL(url).

In a href tag, use this would be worked:

<div ><a href="http://www.ParentShack.com">ParentShack.com</a></div>

If you want to open a new url, it would be worked to use 'window.open(url, "NewWindow")' ? No. You should use this one:

self.location.href=url;

And don't forget to use menu.clear() to clear the menu before open a URL.