Displaying the titles of an RSS feed: Difference between revisions
(Created page with "<source lang="python"> #!/usr/bin/env python from optparse import OptionParser parser = OptionParser() parser.add_option("-u", "--url", dest="url", default="http://feeds.bbci.co...") |
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Update: See [[!Category:argparse]] | |||
<source lang="python"> | <source lang="python"> | ||
#!/usr/bin/env python | #!/usr/bin/env python | ||
Revision as of 13:37, 26 October 2011
Update: See !Category:argparse
#!/usr/bin/env python
from optparse import OptionParser
parser = OptionParser()
parser.add_option("-u", "--url", dest="url", default="http://feeds.bbci.co.uk/news/rss.xml", help="the url to read from")
parser.add_option("-n", "--numlines", type="int", dest="num", default=1000, help="how many lines to display")
(options, args) = parser.parse_args()
import feedparser
feed = feedparser.parse(options.url)
for e in feed.entries[:options.num]:
print e.title.encode("utf-8")
#!/usr/bin/env python
from optparse import OptionParser
parser = OptionParser()
parser.add_option("-u", "--url", dest="url", default="http://feeds.bbci.co.uk/news/rss.xml", help="the url to read from")
parser.add_option("-n", "--numlines", type="int", dest="num", default=1000, help="how many lines to display")
(options, args) = parser.parse_args()
import feedparser
feed = feedparser.parse(options.url)
for e in feed.entries[:options.num]:
print e.title.encode("utf-8")