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

[iain]

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.

[iain]

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

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.

[dkarp]

Try something like "zmmailbox gru ?id=<your-msg-id>"

[iain]

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>

[dkarp]

To collapse steps 1 and 2, you can use the -t message option to zmmailbox search to search for messages instead of conversations.

[iain]

Good tip - thanks.

s -t message in:Inbox

does the trick nicely.