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Thread: [SOLVED] How to use zmmailbox to "show original"

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  1. #1
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    Default [SOLVED] How to use zmmailbox to "show original"

    Is there a way to view the raw ("show original") source of an email from the command line?

    I've tried using the getMessage subcommand of zmmailbox, but that doesn't show all the headers.

    Thanks,

    Iain.

  2. #2
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    Default

    As a partial answer, using the -v flag with getMessage (or gm) shows some of the header information. It misses out routing information. However, the command zmmsgtrace gives that information.

    So for example

    Code:
    zmmailbox
    mbox> aa admin@myserver.com password
    mbox> sm user@myserver.com
    mbox> gaf
    mbox> gf Inbox
    mbox> s in:Inbox
    mbox> gm -v 1274
    mbox> quit
    
    zmmsgtrace -s user@hotmail.com -r user@myserver.com
    Having said that, I'm still trying to figure out how the message IDs work: when referencing a message by ID, it sometimes says it isn't there with one zmmailbox subcommand whilst appearing to show up with another.

  3. #3
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    Default Get it via REST

    Try something like "zmmailbox gru ?id=<your-msg-id>"
    Bugzilla - Wiki - Downloads - Before posting... Search!

  4. #4
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    Default

    Thanks, that did the trick.

    Having logged onto zmmailbox as admin and opened the mailbox to check, it seems to be a three-step process:

    1. Find the conversation
    e.g.
    s "in:Inbox zimbra"

    2. From the conversation, find the message
    gc <conversation id>

    3. View the message raw text
    gru ?id=<message id>
    Last edited by iain; 09-02-2007 at 03:07 PM.

  5. #5
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    Default -t message

    To collapse steps 1 and 2, you can use the -t message option to zmmailbox search to search for messages instead of conversations.
    Bugzilla - Wiki - Downloads - Before posting... Search!

  6. #6
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    Default

    Good tip - thanks.

    s -t message in:Inbox

    does the trick nicely.

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