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What is the significance of sh -s in ssh -qtt ${user}@${host} "sh -s "${version}"" < test.sh?


 
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Top Forums UNIX for Dummies Questions & Answers What is the significance of sh -s in ssh -qtt ${user}@${host} "sh -s "${version}"" < test.sh?
# 1  
Old 08-15-2013
What is the significance of sh -s in ssh -qtt ${user}@${host} "sh -s "${version}"" < test.sh?

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


Last edited by Franklin52; 08-16-2013 at 03:21 AM.. Reason: Please use code tags
# 2  
Old 08-15-2013
it'd be used by test.sh. similar to if you were to do, ./test.sh "$version"
# 3  
Old 08-15-2013
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""

# 4  
Old 08-15-2013
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
# 5  
Old 08-16-2013
Quote:
Originally Posted by Sree10
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)
 

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