UNIX for Dummies Questions & Answers

Please can you help me understand the significance of providing arguments under sh -s in

Code:

> ssh -qtt ${user}@${host} "sh -s "${version}"" < test.sh

it'd be used by test.sh. similar to if you were to do, ./test.sh "$version"

Thanks!, but please can you explain me how exactly that happens with

Code:

ssh -qtt $user@$host "sh -s "$version"" < test.sh

and not with

Code:

ssh -qtt $user@$host  < "test.sh "$version""

If you run your second ssh command, sh (or better: that user's default shell from /etc/passwd) is implicitely run on the remote system, test.sh being fed into stdin. No process/command invocation - no parameters.
In the first version, you explicitely invoke sh and thus can supply parameters; the -s option tells it to then read commands from stdin. (actually, I'm not sure if two shs run, one default shell which then invokes sh with options and parameters)

From man sh:

Code:

-s stdin         Read commands from standard input (set automatically if no file arguments are present).

Also, check this link and the examples provided therein: Bourne Shell shell-session option