Ssh to an array of servers in a for loop

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# 1  
Old 09-11-2013
Ssh to an array of servers in a for loop

There are 4 remote hosts that I have stored in an array. A ssh trust has been created from the local host to each of the remote hosts.
I am trying to ssh to each of the servers in a for loop as shown below.

declare -a host

for host in "${host[@]}"
  ssh -qtt user@${host} "sh -s "${version}""  <
done is a file on the local server that I am running on the remote host after I ssh to the remote host. But it is terminating after the first iteration. If I use -n switch it prompts me for password in the second iteration.

Please can you guide me on this issue.


Last edited by Sree10; 09-12-2013 at 07:12 AM..
# 2  
Old 09-11-2013
You cannot double quote double quotes. You could escape them with \, or just use a single quote.

ssh -qtt user@${host} "sh -s '${version}'"  <

# 3  
Old 09-11-2013
for host in "${host[@]}"
ssh -qtt ohsadm@${host} "sh -s '${version}'"  <

exits after the first iteration. host array has 4 servers. I want the ssh in the for loop to run through all the 4 servers.
# 4  
Old 09-11-2013
Try with -n -o BatchMode=yes options:
-n for manage stdin
-o BatchMode=yes for no ask password

# 5  
Old 09-11-2013
You do not want -n when running a script read from stdin! This is not the usual stdin problem.

Can you show your complete, unmodified script? I think something funny is going on here.
# 6  
Old 09-12-2013
I am able to iterate successfully through the array if I specify the commands as mentioned below.

for host in "${host[@]}"
  echo $host
  ssh -qtt user@${host} "cd /build && cp -p web.zip_${version} && rm -rf test && unzip /build/ && cd /bin ";  done;

However, if I try to add the commands in a separate file called and stream it to ssh as shown below, it exits after the first iteration. Pressing Ctrl-C manually triggers the next iteration.

for host in "${host[@]}"
  ssh -qtt user@${host} "sh -s "${version}""  <

# 7  
Old 09-12-2013
I repeat: You cannot put double quotes inside double quotes that way.

This is wrong:

"sh -s "${version}""

This would be more correct:

"sh -s '${version}'"

I just realized, however -- your probably uses the ${version} variable, which is why you are feeding it into sh, yes? You want it to carry over. But arguments to sh -s don't work that way, you don't get the names, just $1 $2 $3. Not to mention, you need -- if you want it to assume the following are arguments. Try this:

$ sh -s -- a b c
$ echo $1


$ echo $2


$ echo $3


$ exit

So I think you actually want:
ssh -qtt user@${host} "sh -s -- '${version}'"  <

...and your script should be using $1 and not ${version} .

I suspect you can omit the -tt completely, too. I see no commands in there that actually need a terminal, and throwing one into the mix when not needed seldom helps.

Last edited by Corona688; 09-12-2013 at 02:25 PM..
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