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dimones / feedparser

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SAXParseException on valid (w3c) rss feed #411

Open GoogleCodeExporter opened 9 years ago

GoogleCodeExporter commented 9 years ago 
What steps will reproduce the problem?
1. feedparser.parse("digitalfernsehen.de/rss/rss91.xml")
2. See SAXParseException('not well-formed (invalid token)')
3. Check 
http://validator.w3.org/feed/check.cgi?url=digitalfernsehen.de%2Frss%2Frss91.xml
 for valid feed status

What is the expected output? What do you see instead?
Expected ~100 stories with short summaries. Saw exception.

What version of the product are you using? On what operating system?
Using Mac and Ubuntu latest. Python 2.7.
>>> feedparser.__version__
'5.1.2'

Please provide any additional information below.
This is to fix a medium-sized feed in terms of popularity. See 
http://www.newsblur.com/site/22353/digital-fernsehen.

Original issue reported on code.google.com by sam...@newsblur.com on 30 Jul 2013 at 10:12

GoogleCodeExporter commented 9 years ago 
Are you still having this problem with the latest code in the git repo? Also, 
can you reproduce this problem if you add "http://" to the front of the URL? 
feedparser may be trying to access that as a file on your local filesystem.

Original comment by kurtmckee on 10 Jul 2014 at 4:56

GoogleCodeExporter commented 9 years ago 
Looks good to me! Thanks for the fix. I merged in v5.1.3 and it works.

Original comment by conesus on 10 Jul 2014 at 5:34