Are you tyring to search for a feed or load a feed? Your first snippet liiked like a search, where this one looks like maybe your trying to lead a feed. The are slightly different urls. /feed/load if loading.
On Mar 17, 9:24 am, crazywizard <[email protected]> wrote: > Adam, I made the change in the url but am still getting an error > code: > > url = ('https://ajax.googleapis.com/ajax/services/feed/find?'+ > 'v=1.0&q=http://www.nation.co.ke/-/1148/1148/-/view/asFeed/-/ > vtvnjq/-/index.xml&key=ABQIAAAA3yxjuwRO-TvCe0atz-bXURSxacz9-9- > HUriveZcWWM1mnHlsLBS0fBq1eNV8LD9XYfb23n6t83I8Tw&userip=192.168.0.1') > > Output: > [root@localhost Desktop]# python feed.py > {'responseData': {'query': 'http://www.nation.co.ke/-/1148/1148/-/view/ > asFeed/-/vtvnjq/-/index.xml', 'entries': []}, 'responseDetails': > 'invalid query', 'responseStatus': 200} -- You received this message because you are subscribed to the Google Groups "Google AJAX APIs" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/google-ajax-search-api?hl=en.
