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I am developing a simple wrapper script for SSH to do some preprocessing before actually executing SSH (to be exact: load the per-host key with keychain/ssh-agent).

In order to do so, I need to extract the host name from the command line arguments. The wrapper shall support all options accepted by SSH — so my first, simple approch using the last command line parameter does not work if the user intends to run a command non-interactively on the remote side.

Does anybody see a simple shell-only way to get the hostname reliably?

Thanks in advance!

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  • As a first pass, I would use getopt to loop through the ssh arguments. The first non-option parameter should be either hostname or user@hostname (where hostname is a name, an IP address, or an alias from the ssh configuration file[s]). Commented Jul 9, 2017 at 12:45
  • $(hostname) ? Commented Jul 9, 2017 at 12:46
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    @user4556274: I thought about this too, but wouldn't I then need to skip arguments to the options (such as for -o)? This would actually mean to name all SSH option in the script, right? Commented Jul 9, 2017 at 16:31
  • @DavidPostill: I'm looking for specifierd remote hostname, not the local one... Commented Jul 9, 2017 at 16:31

1 Answer 1

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Just use ssh -G:

ssh -G -nNT -L10000:localhost:10000 admin@yo|awk '/^hostname/ {print $2}'

Will print yo. It will fully resolve to the real hostname if you have a match in your .ssh/config.

Edit: awk improvement suggested by Grant McLean

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  • Thank you for this - it was exactly what I needed. Incidentally, awk already has built-in grep-like functionality. Your command can be simplified to: ssh -G -nNT -L10000:localhost:10000 admin@yo | awk '/^hostname/ {print $2}' Commented Oct 16, 2024 at 0:58
  • Correct I will fix the answer. Commented Oct 16, 2024 at 20:23

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