I have never worked with an RSS feed before, I can't seem to find the url of the feed.
The page which is offering the RSS Feed:
https://www.sec.gov/edgar/browse/?CIK=717826&owner=exclude
I am using feedparser:
import feedparser
rss_url = 'https://www.sec.gov/edgar/browse/?CIK=717826/.rss'
Feed = feedparser.parse(rss_url)
pointer = Feed.entries[1]
# result is empty
I think I am using the wrong link and can't seem to find the right one. I tried to view source on the RSS button and didn't find a link. The button downloads an XML file when I click it.
Can someone help me understand how to find this link?
The link on RSS button is correct
And the behaviour that you are getting XML document when you go there, is also correct, because RSS is based on XML format, so what
feedparserlibrary is dealing with is actual XML content. It parses it and allow you access the results via Python API.For example, on the page
you have third row with
and in RSS feed (XML format) you have this entry as well:
UPDATE:
On the other hand when you change your code to use the RSS button URL
you will see that the site is blocking your request was blocked:
To adjust that take a look on the documentation development section and in particular programmatic access. You have to use proper
User-Agent: