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Old 04-06-2012, 10:32 AM   #1
cram1010
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Error "Could not fetch link..." while creating a basic recipe

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!
Attached Files
File Type: txt rebelion.recipe.txt (1.0 KB, 53 views)
File Type: txt output.txt (99.7 KB, 56 views)
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Old 05-03-2012, 07:18 PM   #2
atlantique
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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!
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Old 05-04-2012, 11:13 AM   #3
cram1010
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Thanks a lot atlantique. With your tip I got to build the rebelion.org recipe. I have posted it here: http://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:

Code:
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
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Old 05-05-2012, 06:15 AM   #4
vietchovui
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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.
Attached Files
File Type: txt cailbre - TuanVietNam_25.recipe.txt (245 Bytes, 37 views)

Last edited by vietchovui; 05-05-2012 at 11:26 AM. Reason: find out the bug
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Old 05-05-2012, 07:13 AM   #5
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[deleted] please read the above post.

Last edited by vietchovui; 05-05-2012 at 11:28 AM.
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