Re: RSS feed parser
[EMAIL PROTECTED] wrote: On Apr 2, 10:20 pm, Florian Lindner [EMAIL PROTECTED] wrote: Some of the question I have but found answered nowhere: I have a feedparser object that was created from a string. How can I trigger a update (from a new string) but the feedparser should treat the new string like the same feed (thus setting feed.updated etc.). Hmm. Do you mean that the feed object should stay the same? Like the difference between a = [1,2,3]; a = [1,2,3]+[4] and a = [1,2,3]; a.append(4)? I glanced at the parse function in the source code and it looks like it's not directly possible. You could modify it so that the result dictionary is optionally given as an argument, so when updating you'd do: feedparser.parse(string, oldFeed). You'd also have to clear the oldFeed object before update. But you might also be able to solve the problem by using an additional layer of indirection. Instead of passing around the feed object, you'd pass around a proxy object like this: class Empty: pass proxy = Empty() proxy.feed = feedparser.parse(string) storeProxyForLaterUse(proxy) proxy.feed = feedparser.parse(string2) useStoredProxy() #this would use the updated feed through the proxy Then just use proxy.feed.updated everywhere instead of directly feed.updated. A smarter proxy would automatically translate proxy.updated into proxy.feed.updated so usage would stay as simple as without the proxy. Doing this is quite easy in Python (search for __getattr__ examples). I already use something like that (with __getattr__). The problem is that with this way there is still a new feed object created everytime a new string needs to be passed to. But since I want to use use the updated_parsed etc. function it's not possible that each time the feed is parsed a new object is created (the update times will always be the time of the last parsing). Any idea? Regards, Florian -- http://mail.python.org/mailman/listinfo/python-list
RSS feed parser
Hello, I'm looking for python RSS feed parser library. Feedparser http://feedparser.org/ does not seem to maintained anymore. What alternatives are recommendable? Thanks, Florian -- http://mail.python.org/mailman/listinfo/python-list
Re: RSS feed parser
On Apr 2, 7:22 pm, Florian Lindner [EMAIL PROTECTED] wrote: Hello, I'm looking for python RSS feed parser library. Feedparserhttp://feedparser.org/does not seem to maintained anymore. What alternatives are recommendable? Thanks, Florian Well, even if it's not maintained anymore (where does it say that?), it works fine and the API is great. Although of course I do realize that when a new version of RSS appears, feedparser won't be able to support it unless someone updates it. But RSS 2.0 appeared already in 2002 and no new versions have come since. So I wouldn't worry too much about new RSSs popping up every month. Maybe the feedparser code hasn't been updated in a while because it's already perfect and there's nothing to add to it?-) -- http://mail.python.org/mailman/listinfo/python-list
Re: RSS feed parser
[EMAIL PROTECTED] wrote: On Apr 2, 7:22 pm, Florian Lindner [EMAIL PROTECTED] wrote: Hello, I'm looking for python RSS feed parser library. Feedparserhttp://feedparser.org/does not seem to maintained anymore. What alternatives are recommendable? Thanks, Florian Well, even if it's not maintained anymore (where does it say that?), it works fine and the API is great. Although of course I do realize that when a new version of RSS appears, feedparser won't be able to support it unless someone updates it. But RSS 2.0 appeared already in 2002 and no new versions have come since. So I wouldn't worry too much about new RSSs popping up every month. Maybe the feedparser code hasn't been updated in a while because it's already perfect and there's nothing to add to it?-) No postings neither on the mailinglists nor in the forums are being answered. Somewhere he stated that he had turned to another hobby. Some of the question I have but found answered nowhere: I have a feedparser object that was created from a string. How can I trigger a update (from a new string) but the feedparser should treat the new string like the same feed (thus setting feed.updated etc.). - How can I trigger a update from a new file? - Does feedparser has the desired behavior? Regards, Florian -- http://mail.python.org/mailman/listinfo/python-list
Re: RSS feed parser
On Apr 2, 10:20 pm, Florian Lindner [EMAIL PROTECTED] wrote: Some of the question I have but found answered nowhere: I have a feedparser object that was created from a string. How can I trigger a update (from a new string) but the feedparser should treat the new string like the same feed (thus setting feed.updated etc.). Hmm. Do you mean that the feed object should stay the same? Like the difference between a = [1,2,3]; a = [1,2,3]+[4] and a = [1,2,3]; a.append(4)? I glanced at the parse function in the source code and it looks like it's not directly possible. You could modify it so that the result dictionary is optionally given as an argument, so when updating you'd do: feedparser.parse(string, oldFeed). You'd also have to clear the oldFeed object before update. But you might also be able to solve the problem by using an additional layer of indirection. Instead of passing around the feed object, you'd pass around a proxy object like this: class Empty: pass proxy = Empty() proxy.feed = feedparser.parse(string) storeProxyForLaterUse(proxy) proxy.feed = feedparser.parse(string2) useStoredProxy() #this would use the updated feed through the proxy Then just use proxy.feed.updated everywhere instead of directly feed.updated. A smarter proxy would automatically translate proxy.updated into proxy.feed.updated so usage would stay as simple as without the proxy. Doing this is quite easy in Python (search for __getattr__ examples). -- http://mail.python.org/mailman/listinfo/python-list
