Parsing XML RSS feed byte stream for <item> tag

[darrel.rendell]

I'm attempting to parse an RSS feed for the first instance of an element ""..

def pageReader(url):
try:
readPage = urllib2.urlopen(url)
except urllib2.URLError, e:
# print 'We failed to reach a server.'
# print 'Reason: ', e.reason
return 404
except urllib2.HTTPError, e:
# print('The server couldn\'t fulfill the request.')
# print('Error code: ', e.code)
return 404
else:
outputPage = readPage.read()
return outputPage

Assume arguments being passed are correct. The function returns a str object whose value is simply an entire rss feed - I've confirmed the type with:

a = isinstance(value, str)
if not a:
return -1

So, an entire rss feed has been returned from the function call, it's this point I hit a brick wall - I've tried parsing the feed with BeautifulSoup, lxml and various other libs, but no success (I had some success with BeautifulSoup, but it wasn't able to pull certain child elements from the parent,for example, . I'm just about ready to resort to writing my own parser, but I'd like to know if anybody has any suggestions.

To recreate my error, simply call the above function with an argument similar to:

http://www.cert.org/nav/cert_announcements.rss

You'll see I'm trying to return the first child.

<item>
<title>New Blog Entry: Common Sense Guide to Mitigating Insider Threats - Best Practice 16 (of 19)</title>
<link>http://www.cert.org/blogs/insider_t...insider_threats_-_best_practice_16_of_19.html</link>
<description>This sixteenth of 19 blog posts about the fourth edition of the Common Sense Guide to Mitigating Insider Threats describes Practice 16evelop a formalized insider threat program.</description>
<pubDate>Wed, 06 Feb 2013 06:38:07 -0500</pubDate>
</item>

As I've said, BeautifulSoup fails to find both pubDate and Link, which are crucial to my app.

Any advice would be greatly appreciated.

 

[John Gordon]

def pageReader(url):
try:
readPage =3D urllib2.urlopen(url)
except urllib2.URLError, e:
# print 'We failed to reach a server.'
# print 'Reason: ', e.reason
return 404 =20
except urllib2.HTTPError, e:
# print('The server couldn\'t fulfill the request.')
# print('Error code: ', e.code) =20
return 404 =20
else:
outputPage =3D readPage.read() =20
return outputPage

To recreate my error, simply call the above function with an argument
similar to:

You'll see I'm trying to return the first child.

The above code produces no output at all. The pageReader() function is
defined but never called.

If we add a few lines at the bottom:

if __name__ == '__main__':
print pageReader('http://www.cert.org/nav/cert_announcements.rss')

Then we get some output:

<?xml version="1.0" encoding="UTF-8"?>
<rss version="2.0">

<channel>
<title>CERT Announcements</title>
<link>http://www.cert.org/nav/whatsnew.html</link>
<language>en-us</language>
<description>Announcements: What's New on the CERT web site</description>

<item>
<title>New Blog Entry: Common Sense Guide to Mitigating Insider Threats - Best Practice 16 (of 19)</title>
<link>http://www.cert.org/blogs/insider_t...insider_threats_-_best_practice_16_of_19.html</link>
<description>This sixteenth of 19 blog posts about the fourth edition of the Common Sense Guide to Mitigating Insider Threats describes Practice 16: Develop a formalized insider threat program.</description>
<pubDate>Wed, 06 Feb 2013 06:38:07 -0500</pubDate>
</item>

....

As I've said, BeautifulSoup fails to find both pubDate and Link, which are =
crucial to my app.

Any advice would be greatly appreciated.

You haven't included the BeautifulSoup code which attempts to parse the XML,
so it's impossible to say exactly what the error is.

However, I have a guess: you said you're trying to return the first
child. Based on the above output, the first child is the <channel>
element, not an <item> element. Perhaps that's the issue?

 

[xDog Walker]

As I've said, BeautifulSoup fails to find both pubDate and Link, which are
crucial to my app
Any advice would be greatly appreciated.

http://packages.python.org/feedparser