not sure if I'm getting correct response. I might be but don't know how
output is supposed to appear. Please let me know if it already correct but
I don't realize it.
here is complete controller:
def feed():
response.view = 'generic.rss'
return dict(title="my feed", link="http://feed.example.com";,
description="my first feed", entries=[dict(title="my feed",
link="http://feed.example.com";, description="my first feed")])
there is new view for it
result of URL http://127.0.0.1:8000/ES3/default/feed.rss is
<?xml version="1.0" encoding="utf8"?>
<rss version="2.0"><channel><title>my
feed</title><link>http://feed.example.com</link><description>my first
feed</description><lastBuildDate>Fri, 25 Nov 2016 22:52:08
GMT</lastBuildDate><generator>PyRSS2Gen-1.1.0</generator><docs>http://blogs.law.harvard.edu/tech/rss</docs><item><title>my
feed</title><link>http://feed.example.com</link><description>my first
feed</description><pubDate>Fri, 25 Nov 2016 22:52:08
GMT</pubDate></item></channel></rss>
result of URL http://127.0.0.1:8000/ES3/default/feed is
http://feed.example.commy first feedFri, 25 Nov 2016 22:51:07
GMTPyRSS2Gen-1.1.0http://blogs.law.harvard.edu/tech/rsshttp://feed.example.commy
first feedFri, 25 Nov 2016 22:51:07 GMT
that may be correct but I thought there is a clickable link that the rss
creates that takes user to the real site updated content html page
thanks for helping me on Thanksgiving holiday
--
Resources:
- http://web2py.com
- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
- https://code.google.com/p/web2py/issues/list (Report Issues)
---
You received this message because you are subscribed to the Google Groups
"web2py-users" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to web2py+unsubscr...@googlegroups.com.
For more options, visit https://groups.google.com/d/optout.
