Hello all,
I am trying to execute in a shell script where, a and b are variables
a=username and b=server.ip
It is giving me an error why am I getting the error?? ssh does not accept variables? Can anyone help.
Thanks in Advance

Try putting single quote around the command you want ssh to execute @b and also put brackets around your variables to ensure it sees them distinctly:
ssh ${a}@${b} 'mkdir abc'

Also if that still doesn't work be sure to verify a simple ssh works as you expect:
ssh <user>@<hostIP>
It's always possible your issue has nothing to do with variables but simply a failure to connect to the remote server as the user you're specifying which is what the error is telling you. Command line without variables would give same error if that is the case.

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thanks... Yeah I have tried using the basic ssh <user>@<ip> command and it works perfectly.. will try the ssh ${a}@${b} 'mkdir abc' command and let u know abt the result..

What shell are you using?
In bash, neither the braces (since variable names cannot contain '@') nor the single quotes are necessary, although the quotes are recommended and the braces make the command more readable. The command you showed in the original post just has to work the way it is. Please check the contents of the variables a and b and make sure you didn't misspell any of them.

tested : ssh ${a}@${b} 'mkdir -p .ssh' but its not working

i am executing .sh file from cygwin terminal

contents: #!/bin/bash
a=`grep -Po '(?<=Username=).*' /cygdrive/d/vj/rs`
b=`grep -Po '(?<=Password=).*' /cygdrive/d/vj/rs`
c=`grep -Po '(?<=IP ADD=).*' /cygdrive/d/vj/rs`
echo -ne '\n' | ssh-keygen -t rsa -N "" -f id_rsa
echo -ne '\n'
echo -ne '\n'
echo -ne '\n' | ssh ${a}@${b} 'mkdir -p .ssh'

manually, all commands are working perfectly.
automatically, all commands are working except last command.
its not accepting '$' (maually & automatically)
giving an error: ssh: connect to host port 22: Connection refused

please use [code]here comes your script[/code] to keep formatting
also try
to see what's happening. (That will not solve anything, just print the lines as they executed, so you will be able to check tthc program flow).
Also at the last line you do not need echo:
can you please explain what do you mean by "its not accepting '$' (maually & automatically)"

in the above.. a, b ,c is a variable in which values of username, password, IP address stored from the file (rs).
checked these three commands one by one maually.. its working
if i write: ssh $a@$b 'mkdir -p ~/.ssh' instead of ssh root@XXX.XXX.XXX.XXX 'mkdir -p ~/.ssh'
then it is not taking value of a and b

Yeah - but if "b" is supposed to be the password it definitely won't work.

sorry typing mistake..
its ssh $a@$c 'mkdir -p ~/.ssh'
where a is 'root' and c is IP Address

Now add the "-vx" option to the bash command and see what is happening.

what you are looking for is:

echo 'mkdir abcd' | sshpass -p 'MyPassWord' ssh root@1.2.3.4 bash

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You don't need (and likely don't want) the echo. Just using "ssh user@host mkdir abcd" is enough.

True, though I actually replaced cat with echo from a much larger routine. A slice of that routine would look like:

cat MyScript |sshpass -p $PASSWORD ssh root@$SERVERLIST bash

But the script needed each line to end in ;\ for it to send correctly.

Just as an asideIs meaninless, the -n switch tells echo to suppress the trailing newline and the -e switch tells it to accept escape codes, so you are saying efectivly don't use new line and add new line, the above command is just the same an echo with no arguments.

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In which case I would expect that
Would work better.

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