Grokbase › Groups › Python › mailman-users › February 2009
FAQ
I was wondering if anyone knows of any command you can use to find out
who is subscribed to all mailman lists.

Similar to the which <email address> command in majordomo?

Thanks in advance.


Brian Canty
Manager Computer Information Services
American Psychoanalytic Association
212-752-0450 x17

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  • Mark Sapiro at Feb 10, 2009 at 5:19 pm ⇧

    Brian Canty wrote:
    I was wondering if anyone knows of any command you can use to find out
    who is subscribed to all mailman lists.

    Similar to the which <email address> command in majordomo?

    If you have command line access, see

    bin/find-member --help

    A user (but not a list admin) can go to the web user options page and
    "List my other subscriptions".

    There is no email command for this.

    --
    Mark Sapiro <mark at msapiro.net> The highway is for gamblers,
    San Francisco Bay Area, California better use your sense - B. Dylan
  • Karl O. Pinc at Feb 10, 2009 at 8:07 pm ⇧

    On 02/10/2009 10:20:51 AM, Brian Canty wrote:
    I was wondering if anyone knows of any command you can use to find out
    who is subscribed to all mailman lists.
    list_lists | tail -n +2 | awk '{print $1;}' | xargs -n 1 list_members |
    sort -u

    Gets you all the members of all the lists.

    list_lists \
    tail -n +2 \
    awk '{print $1;}' \
    xargs -n 1 bash -c 'list_members $0 | xargs -n 1 echo $0:' \
    grep foo at example.com \
    cut -d f 1
    Gets you all the lists that foo at example.com is subscribed to.


    Karl <kop at meme.com>
    Free Software: "You don't pay back, you pay forward."
    -- Robert A. Heinlein
  • Karl O. Pinc at Feb 11, 2009 at 2:41 pm ⇧

    On 02/10/2009 02:07:23 PM, Karl O. Pinc wrote:
    On 02/10/2009 10:20:51 AM, Brian Canty wrote:
    I was wondering if anyone knows of any command you can use to find
    out
    who is subscribed to all mailman lists.
    Somehow the "cut" command got mangled. Should be:

    list_lists \
    tail -n +2 \
    awk '{print $1;}' \
    xargs -n 1 bash -c 'list_members $0 | xargs -n 1 echo $0:' \
    grep foo at example.com \
    cut -d : -f 1
    Gets you all the lists that foo at example.com is subscribed to.

    Karl <kop at meme.com>
    Free Software: "You don't pay back, you pay forward."
    -- Robert A. Heinlein
  • Mark Sapiro at Feb 11, 2009 at 4:40 pm ⇧

    Karl O. Pinc wrote:
    On 02/10/2009 02:07:23 PM, Karl O. Pinc wrote:
    On 02/10/2009 10:20:51 AM, Brian Canty wrote:
    I was wondering if anyone knows of any command you can use to find
    out
    who is subscribed to all mailman lists.
    Somehow the "cut" command got mangled. Should be:

    list_lists \
    tail -n +2 \
    awk '{print $1;}' \
    xargs -n 1 bash -c 'list_members $0 | xargs -n 1 echo $0:' \
    grep foo at example.com \
    cut -d : -f 1
    Gets you all the lists that foo at example.com is subscribed to.

    So does

    bin/find_member foo at example.com

    and much more simply.

    --
    Mark Sapiro <mark at msapiro.net> The highway is for gamblers,
    San Francisco Bay Area, California better use your sense - B. Dylan
  • Karl O. Pinc at Feb 12, 2009 at 4:49 am ⇧

    On 02/11/2009 10:40:50 AM, Mark Sapiro wrote:
    Karl O. Pinc wrote:
    Gets you all the lists that foo at example.com is subscribed to.

    So does

    bin/find_member foo at example.com

    and much more simply.
    And I'm sure more efficiently too. My bad.

    Karl <kop at meme.com>
    Free Software: "You don't pay back, you pay forward."
    -- Robert A. Heinlein
  • Mark Sapiro at Feb 11, 2009 at 4:43 pm ⇧

    Karl O. Pinc wrote:
    On 02/10/2009 10:20:51 AM, Brian Canty wrote:
    I was wondering if anyone knows of any command you can use to find out
    who is subscribed to all mailman lists.
    list_lists | tail -n +2 | awk '{print $1;}' | xargs -n 1 list_members |
    sort -u

    Gets you all the members of all the lists.

    So does

    bin/find_member .

    or

    bin/find_member @

    --
    Mark Sapiro <mark at msapiro.net> The highway is for gamblers,
    San Francisco Bay Area, California better use your sense - B. Dylan

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groupmailman-users @
categoriespython
postedFeb 10, '09 at 4:20p
activeFeb 12, '09 at 4:49a
posts7
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3 users in discussion

Mark Sapiro: 3 posts Karl O. Pinc: 3 posts Brian Canty: 1 post

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