KeyError 'url' when trying to add feed.

[DonkeyShoot]

When trying to add a feed, I receive a keyError: 'url' error.

Stacktrace:

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/flask/app.py", line 1997, in __call__

    return self.wsgi_app(environ, start_response)

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/tembozapp/server.py", line 50, in __call__

    return self.application(environ, start_response)

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/flask/app.py", line 1985, in wsgi_app

    response = self.handle_exception(e)

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/flask/app.py", line 1540, in handle_exception

    reraise(exc_type, exc_value, tb)

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/flask/app.py", line 1982, in wsgi_app

    response = self.full_dispatch_request()

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/flask/app.py", line 1614, in full_dispatch_request

    rv = self.handle_user_exception(e)

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/flask/app.py", line 1517, in handle_user_exception

    reraise(exc_type, exc_value, tb)

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/flask/app.py", line 1612, in full_dispatch_request

    rv = self.dispatch_request()

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/flask/app.py", line 1598, in dispatch_request

    return self.view_functions[rule.endpoint](**req.view_args)

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/tembozapp/server.py", line 546, in add_feed

    = update.add_feed(feed_xml)

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/tembozapp/update.py", line 115, in add_feed

    'xmlUrl': f['url'],

    File "/root/Projects/temboz/pythonenv/tembozdir/lib/python2.7/site-packages/tembozapp/feedparser.py", line 374, in __getitem__

    return dict.__getitem__(self, key)

    KeyError: 'url'

I've tried this with multiple feeds, including feeds from the me.opml file

[fazalmajid]

If you don’t mind sharing, what’s the feed URL you were trying to subscribe to?

[DonkeyShoot]

Dear Fazal Majid,

Sorry for my late response on this issue. I was trying several urls, e.g. : https://www.schneier.com/blog/atom.xml https://www.eff.org/rss/updates.xml https://blog.qualys.com/feed

I also tried a feed that was present in the me.opml file, namely: http://lenspire.zeiss.com/en/feed/

All of them triggered the error.

Thanks a lot for your response!

KR, Don

2017-10-10 16:51 GMT+02:00 Fazal Majid notifications@github.com:

If you don’t mind sharing, what’s the feed URL you were trying to subscribe to?

— You are receiving this because you authored the thread. Reply to this email directly, view it on GitHub https://github.com/fazalmajid/temboz/issues/117#issuecomment-335498656, or mute the thread https://github.com/notifications/unsubscribe-auth/AYQnLv4lrKfytMqTeGWxt46GX3XSclAUks5sq4RfgaJpZM4Pzxgq .

[fazalmajid]

I am assuming you followed the instructions and installed using virtualenv/pip. I hadn't updated the package on PyPI for a while, and clearly there were bugs in it that have since been fixed in git. I pushed a new version 2.1.0 of temboz to PyPI.

I reproduced your problem on my Mac using Temboz 2.0.0 and verified they are fixed in 2.1.0, please give it a try, and thanks for your interest in this software!