Comments (3)
carltongibson
commented on June 6, 2024
1
Works for me:
>>> import feedparser
>>> d = feedparser.parse('https://hackernewsrss.com/feed.xml')
>>> d.keys()
dict_keys(['bozo', 'entries', 'feed', 'headers', 'etag', 'href', 'status', 'encoding', 'version', 'namespaces'])
>>> d["feed"].keys()
dict_keys(['title', 'title_detail', 'subtitle', 'subtitle_detail', 'links', 'link', 'language', 'updated', 'updated_parsed', 'published', 'published_parsed', 'sy_updateperiod', 'image'])
>>> d["feed"]["title"]
'Hacker News RSS Feed'
>>>
from feedparser.
bdim404
commented on June 6, 2024
Ok, I got it! Thanks for your reply!
from feedparser.
Related Issues (20)
- Parser freezes terminal with no error HOT 4
- enclosures property does not exist if entry has an id but no link HOT 1
- the example from the documentation fails with an error HOT 3
- Link w/o href considered alternate unconditionally
- `py.typed` is missing from current release on pypi HOT 2
- Potential security issue HOT 5
- having image tag within items within RSS 2.0 feeds HOT 1
- Crash in feedparser 6.0.10 HOT 6
- Docs: Wrong link to Python Unicode tutorial HOT 3
- FR: With items with no link but an enclosure with a href, use that href for entries[i].link HOT 1
- NASDAQ RSS feed not parsed HOT 2
- 'updated' kludge status
- duplicate elements dropped during parse
- Summary overwritting content
- seems to return empty feed in case of network errors
- Make feedparser accepts caseinsensitive headers
- `io.StringIO` not working in `parse()`
- AssertionError: unexpected '/' char in declaration when parsing feed HOT 1
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from feedparser.