Re: [Freevo-users] Is it a bug in rssserver? (freevo-1.7.4)

2007-11-19 Thread Laurento Frittella 

Il giorno lun, 19/11/2007 alle 21.42 +0100, Duncan Webb ha scritto:
> Since you have posted this as a but too, it is possible that you have a
> configuration error or that there is a bug. You really need to post your
> rss.feeds file.

I posted it in bug report #1834529

Cheers,
Laurento


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Re: [Freevo-users] Is it a bug in rssserver? (freevo-1.7.4)

2007-11-19 Thread Duncan Webb 
Laurento Frittella wrote:
> I'm trying to use rssserver. The first run worked fine, after a reboot I
> receive this error:
> 
> 
> Freevo 1.7.4 r10133 started at Sat Nov 17 22:41:10 2007
> 
> 2007-11-17 22:41:10,614 INFO config.py (477):
> logdir: /var/log /var/log/freevo
> 2007-11-17 22:41:10,615 INFO config.py (478):
> staticdir: /var/lib /var/lib/freevo
> 2007-11-17 22:41:10,617 INFO config.py (479):
> cachedir: /var/cache /var/cache/freevo
> 2007-11-17 22:41:10,619 INFO config.py (498): Loading freevo
> configuration file "/etc/freevo/freevo.conf"
> 2007-11-17 22:41:10,849 INFO config.py (585): Loading local
> configuration file "/etc/freevo/local_conf.py"
> 2007-11-17 22:41:10,886 INFO config.py (983):
> overlaydir: /var/cache/freevo/vfs
> 2007-11-17 22:41:10,909 INFO new process watcher instance
> Traceback (most recent call last):
>   File "/usr/lib/python2.4/site-packages/freevo/helpers/rssserver.py",
> line 91, in ?
> t = threading.Thread(rssperiodic.checkForUpdates())
>   File "/usr/lib/python2.4/site-packages/freevo/rssperiodic.py", line
> 238, in checkForUpdates
> for item in rssfeed.Feed(feedSource).items:
>   File "/usr/lib/python2.4/site-packages/freevo/rssfeed.py", line 42, in
> __init__
> self.parseFeed(inputSource)
>   File "/usr/lib/python2.4/site-packages/freevo/rssfeed.py", line 109,
> in parseFeed
> newItem.type =
> getType(re.split('"',re.split("\.",newItem.url)[-1])[0])
>   File "/usr/lib/python2.4/sre.py", line 157, in split
> return _compile(pattern, 0).split(string, maxsplit)
> TypeError: expected string or buffer

Since you have posted this as a but too, it is possible that you have a
configuration error or that there is a bug. You really need to post your
rss.feeds file.

Duncan


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[Freevo-users] Is it a bug in rssserver? (freevo-1.7.4)

2007-11-17 Thread Laurento Frittella 
I'm trying to use rssserver. The first run worked fine, after a reboot I
receive this error:


Freevo 1.7.4 r10133 started at Sat Nov 17 22:41:10 2007

2007-11-17 22:41:10,614 INFO config.py (477):
logdir: /var/log /var/log/freevo
2007-11-17 22:41:10,615 INFO config.py (478):
staticdir: /var/lib /var/lib/freevo
2007-11-17 22:41:10,617 INFO config.py (479):
cachedir: /var/cache /var/cache/freevo
2007-11-17 22:41:10,619 INFO config.py (498): Loading freevo
configuration file "/etc/freevo/freevo.conf"
2007-11-17 22:41:10,849 INFO config.py (585): Loading local
configuration file "/etc/freevo/local_conf.py"
2007-11-17 22:41:10,886 INFO config.py (983):
overlaydir: /var/cache/freevo/vfs
2007-11-17 22:41:10,909 INFO new process watcher instance
Traceback (most recent call last):
  File "/usr/lib/python2.4/site-packages/freevo/helpers/rssserver.py",
line 91, in ?
t = threading.Thread(rssperiodic.checkForUpdates())
  File "/usr/lib/python2.4/site-packages/freevo/rssperiodic.py", line
238, in checkForUpdates
for item in rssfeed.Feed(feedSource).items:
  File "/usr/lib/python2.4/site-packages/freevo/rssfeed.py", line 42, in
__init__
self.parseFeed(inputSource)
  File "/usr/lib/python2.4/site-packages/freevo/rssfeed.py", line 109,
in parseFeed
newItem.type =
getType(re.split('"',re.split("\.",newItem.url)[-1])[0])
  File "/usr/lib/python2.4/sre.py", line 157, in split
return _compile(pattern, 0).split(string, maxsplit)
TypeError: expected string or buffer



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