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""

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

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)