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support XML feeds without an explicit title

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fixes #24
This commit is contained in:
Kyle Mahan 2015-06-01 08:05:37 -07:00
parent ff5c7feb63
commit cf49cfea2c
1 changed files with 17 additions and 10 deletions
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27
woodwind/views.py
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@ -316,21 +316,28 @@ def subscribe():
def add_subscription(origin, feed_url, type, tags=None):
feed = Feed.query.filter_by(feed=feed_url, type=type).first()
if not feed:
name = None
if type == 'html':
flask.current_app.logger.debug('mf2py parsing %s', feed_url)
parsed = mf2util.interpret_feed(
mf2py.Parser(url=feed_url).to_dict(), feed_url)
name = parsed.get('name')
if not name or len(name) > 140:
p = urllib.parse.urlparse(origin)
name = p.netloc + p.path
feed = Feed(name=name, origin=origin, feed=feed_url, type=type)
elif type == 'xml':
flask.current_app.logger.debug('feedparser parsing %s', feed_url)
parsed = feedparser.parse(feed_url)
feed = Feed(name=parsed.feed and parsed.feed.title,
origin=origin, feed=feed_url, type=type)
if parsed.feed:
name = parsed.feed.get('title')
else:
flask.current_app.logger.error('unknown feed type %s', type)
flask.abort(400)
if not name:
p = urllib.parse.urlparse(origin)
name = p.netloc + p.path
feed = Feed(name=name, origin=origin, feed=feed_url, type=type)
if feed:
db.session.add(feed)
@ -386,10 +393,9 @@ def find_possible_feeds(origin):
})
# look for link="feed"
for furl, fprops in parsed.get('rel-urls', {}).items():
if 'feed' in fprops.get('rels', []) and (
not fprops.get('type')
or fprops.get('type') == 'text/html'):
for furl in parsed.get('rels', {}).get('feed', []):
fprops = parsed.get('rel-urls', {}).get(furl, {})
if not fprops.get('type') or fprops.get('type') == 'text/html':
feeds.append({
'origin': origin,
'feed': furl,
@ -406,6 +412,7 @@ def find_possible_feeds(origin):
'type': 'xml',
'title': link.get('title'),
})
return feeds