Error "Could not fetch link..." while creating a basic recipe

[cram1010]

Hi,

I'm trying to create a very simple recipe for the site http://www.rebelion.org

I'm trying to just feed from one RSS source: http://rebelion.org/rss_portada.php

I don't know why, it's not working. For each article it tries to download, following error like this one is given:

Fetching http://www.rebelion.org/noticias/ame...olonial-147596
Traceback (most recent call last):
File "/usr/lib/calibre/calibre/utils/threadpool.py", line 95, in run
(request, request.callable(*request.args, **request.kwds))
File "/usr/lib/calibre/calibre/web/feeds/news.py", line 811, in fetch_article
return self._fetch_article(url, dir, f, a, num_of_feeds)
File "/usr/lib/calibre/calibre/web/feeds/news.py", line 807, in _fetch_article
raise Exception(_('Could not fetch article. Run with -vv to see the reason'))
Exception: Could not fetch article. Run with -vv to see the reason

I attach two files. The recipe one and the full output I receive when trying to create with -vv. Format is txt to allow uploading.

Thank you!

[atlantique]

The typical format of an url included in one of the rebelion.org rss feeds looks like this:

http://www.rebelion.org/noticias/ee....nicomio-148960

In order to avoid the 404 "Not found" problem, it is necessary to convert the URL
to the following format:

http://www.rebelion.org/noticia.php?...-al-manicomio-

Doing this for each of the url listed on the rss feed will solve your problem. However
I do not know how to do this using calibre/python regular expressions, although
one can presume that it must simple to do it, since it is rather straightforward
using set/awk.

Let us hope that one of the more knowledgeable users will help with this url
modification. Good luck!

[cram1010]

Thanks a lot atlantique. With your tip I got to build the rebelion.org recipe. I have posted it here: https://www.mobileread.com/forums/sho....php?p=2068864

Anyway it's strange, why 404 code is returned if the url do exist?

I explain how I solved the issue:

In fact, I realized the ' titular' parameter is no needed, and correct url can simply be:

http://www.rebelion.org/noticia.php?id=148960

As the id is the last sequence of digits of the incorrect url, we can do:

import re #import regexp module
class RebelionRecipe (BasicNewsRecipe):
  #[...Some code...]
  def print_version(self, url):
     id = re.compile('\d*$').search(url).group() #'\d*$' matches 'last serie of digits in a string', which we search in the url. group() returns matched string, which is the id
     return u'http://www.rebelion.org/noticia.php?id=%s' % id

[vietchovui]

I have prob with this feed: http://tuanvietnam.net/home/rss
The article title in the feed is something like this http://tuanvietnam.vietnamnet.vn/XYZ
but the actual article link is http://tuanvietnam.net/XYZ. So, what can I do to make calibre work with this feed?
Added: I have read about replace/join code in the thread Recipes - Re-usable code. But I still cannot fix this prob myself. I'm completely noob in programming.

[vietchovui]

[deleted] please read the above post.