I'd like to know if it's possible to have gmail automatically mark duplicate issues messages as read. We send issue updates for all NeedsReview, NeedsBetterPatch, and PassedReview issues to the patches list, and in addition, any time I star an issue or am CCed on one, I get an email personally about updates. I am already subscribed to the issues list, so sometimes I get three identical updates for the same issue. Do you know if it's possible to mark all but one as read automatically?
Also, do you know if it's possible to automatically mark as read issue updates for comments that I myself made? I already know what those are, since I wrote them, so I don't need to read them again. Aaron Meurer On Tue, Jun 14, 2011 at 7:26 PM, Ondrej Certik <[email protected]> wrote: > Hi, > > Previously I was filtering the issues change emails by matching a name > "sympy-issues" (I think) in there, which caused all emails, including > if somebode CCed me in the issue, to go into my "sympy-issues" label > in gmail, which currently has 1883 unread emails, as I don't have > enough time to keep up with everything. > > However, couple days ago I figured out, that I can filter the general > changes emails by matching list:"<sympy-issues.googlegroups.com>", and > then all issues changes, that CC me are delivered to my inbox. > > So I think that's the best solution, so that I can contribute to > important issues where somebody CCs me. > > Ondrej > > -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/sympy?hl=en. > > -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
