On 20 March 2014 08:25, Matthew Miller <mattdm@xxxxxxxxxxxxxxxxx> wrote:
One of the frustrations of having multiple different lists is that we can't
really easily cross-post, because if people reply but aren't subscribed to
all lists, they get bounces, and the conversation ends up fragmented. Or the
alternative, multi-posting (same post individually to multiple lists), but
then you're _starting_ with fragmentation.
The radical solution to this would be to use topics, but those require users
to change their habits (putting topics in the subject or a keywords
header).
So instead, could we (now, or in the mm3 future) make it so any address
subscribed to any public fedora list is automatically added to an accept
filter for all other lists?
I don't know enough of the mailman internals to know how this might be
implemented, butit seems like it must be doable. And there might be some
details to be worked out around removing people, and possible moderation.
But what do you think of the general idea?
I would be -1 to it. The main reasons are:
1) The person who has done this is now cut out of the thread on the lists he is not subscribed to.
2) The people who are replying to the thread won't know htye aren't talking with this person again.
3) One of the reasons for subscriber only is to cut down SPAM a little bit. This is by making sure that if you are going to send crap to every list you need to subscribe to them. Now you can subscribe to one and get all the lists as a freebie.
4) There will need to be a ton of meta-logic on which lists this is allowed to and which lists it is not. There are lists that do not want to be cc'd random threads no matter how on topic they might be because it is meant to be a place where only the people on the list can talk (board-private or similar lists).
5) This will be an out of band patch for both mailman 2 and mailman 3 though probably just mailman 3 because I don't see it being a simple drop in because of the rules needed in 4. [Does your list allow cross posts from other lists. Which lists?]
Stephen J Smoogen.
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