03-31-2014
Quote:
Originally Posted by
kumar85shiv
but here the problem is, it is ssh to the first line, i mean first server in the redhat_servers.txt file and then script is not moving forward.
The problem is, ssh tries to read from standard input to get more commands.
What is standard input, inside that loop? "redhat_servers.txt". So the ssh command 'eats' all the lines.
ssh -n, or ssh < /dev/null, fixes this as others have noted.
Good luck
10 More Discussions You Might Find Interesting
1. Programming
Hi,
i have been developing a shell script to transfer a set of files from one ubuntu system to another.
Task: while executing the script the files ( ls, dir, cat) in the source machine should transfer to destination machine(at /home/mac/mac/bin)
While the script is executed once again, It... (0 Replies)
Discussion started by: srijith
0 Replies
2. Shell Programming and Scripting
Hi Guys!
I am trying to write a shell script for automated ssh. vairable user and passwd have initialized correctly, but when I use the following it still prompting me for the password.
#!/usr/bin/bash
user='root@10.14.76.225'
passwd='admin'
ssh $user
$passwd
uptime
exit
I... (3 Replies)
Discussion started by: pinpe
3 Replies
3. Debian
Hi,
I've been looking for a way to execute a console program (is in windows but by now I accept the linux way) from a linux machine, but this program has to be opened in the remote side. Linux machine acts only as a "signaling" host. My program has to open the camera in the remote side, but only... (7 Replies)
Discussion started by: zauberberg
7 Replies
4. Shell Programming and Scripting
Hi, I'm new to C Shell programming. I'm trying to check if a sym link exists on remote server if not send email. I'm not having much luck. Can anyone help?
Here is what I have written but it doesn't work. It tells me that my variable was not defined. Here is part of the script, the second... (0 Replies)
Discussion started by: CDi
0 Replies
5. Shell Programming and Scripting
Hi,
I am using ksh , when i try to use for loop i am getting the expected output.
$for variable in $(ps -fu user | grep -i something/ | grep -i something | grep -v grep | awk '{print $2}');do
> grep $variable /tmp/some_path/*/*
> done
when tried the below to remote server, getting... (4 Replies)
Discussion started by: karthikram
4 Replies
6. Shell Programming and Scripting
I have about 500 hosts where I need to ssh by sending the password on the command line or in a text file in a clear text . However I am not able to download "sshpass" or other tools .
Any other ways to pass the password in a script ? (3 Replies)
Discussion started by: gubbu
3 Replies
7. Shell Programming and Scripting
Hello,
I am trying to login to multiple servers and i have to run multiple loops to gather some details..Could you please help me out.
I am specifically facing issues while running for loops.
I have to run multiple for loops in else condition. but the below code is giving errors in for... (2 Replies)
Discussion started by: mohit_vardhani
2 Replies
8. Shell Programming and Scripting
Dear Folks,
I am trying to read a config file contains ip and port numbers.
i want to read each line of the config file and check ssh connection is happening or not.
Kindly guide.
Config file:
abc@1.2.342 22
abc@1.2.343 22
abc@1.2.344 22
abc@1.2.345 22... (9 Replies)
Discussion started by: sadique.manzar
9 Replies
9. Solaris
Am trying to copy a tar file onto a series of remote hosts and untar it at the destination. Need to do this without having to do multiple ssh.
Actions to perform within a single ssh session via shell script
- copy a file
- untar at destination (remote host)
OS : Linux RHEL6 (3 Replies)
Discussion started by: sankasu
3 Replies
10. Shell Programming and Scripting
Hi all,
i'm trying to gether multiple pattern on remote hosts, and trying to print hostname and the pattern,
ssh remoteserver1 -C 'hostname 2>&1;cat /var/log/server1.log | awk -F ";" '"'"'{ print " "$2" "$5}'"'"'| sort | uniq -c | sort -g -r '
The output is the following,
remoteserver1
... (8 Replies)
Discussion started by: charli1
8 Replies
LEARN ABOUT CENTOS
shell-quote
SHELL-QUOTE(1) User Contributed Perl Documentation SHELL-QUOTE(1)
NAME
shell-quote - quote arguments for safe use, unmodified in a shell command
SYNOPSIS
shell-quote [switch]... arg...
DESCRIPTION
shell-quote lets you pass arbitrary strings through the shell so that they won't be changed by the shell. This lets you process commands
or files with embedded white space or shell globbing characters safely. Here are a few examples.
EXAMPLES
ssh preserving args
When running a remote command with ssh, ssh doesn't preserve the separate arguments it receives. It just joins them with spaces and
passes them to "$SHELL -c". This doesn't work as intended:
ssh host touch 'hi there' # fails
It creates 2 files, hi and there. Instead, do this:
cmd=`shell-quote touch 'hi there'`
ssh host "$cmd"
This gives you just 1 file, hi there.
process find output
It's not ordinarily possible to process an arbitrary list of files output by find with a shell script. Anything you put in $IFS to
split up the output could legitimately be in a file's name. Here's how you can do it using shell-quote:
eval set -- `find -type f -print0 | xargs -0 shell-quote --`
debug shell scripts
shell-quote is better than echo for debugging shell scripts.
debug() {
[ -z "$debug" ] || shell-quote "debug:" "$@"
}
With echo you can't tell the difference between "debug 'foo bar'" and "debug foo bar", but with shell-quote you can.
save a command for later
shell-quote can be used to build up a shell command to run later. Say you want the user to be able to give you switches for a command
you're going to run. If you don't want the switches to be re-evaluated by the shell (which is usually a good idea, else there are
things the user can't pass through), you can do something like this:
user_switches=
while [ $# != 0 ]
do
case x$1 in
x--pass-through)
[ $# -gt 1 ] || die "need an argument for $1"
user_switches="$user_switches "`shell-quote -- "$2"`
shift;;
# process other switches
esac
shift
done
# later
eval "shell-quote some-command $user_switches my args"
OPTIONS
--debug
Turn debugging on.
--help
Show the usage message and die.
--version
Show the version number and exit.
AVAILABILITY
The code is licensed under the GNU GPL. Check http://www.argon.org/~roderick/ or CPAN for updated versions.
AUTHOR
Roderick Schertler <roderick@argon.org>
perl v5.16.3 2010-06-11 SHELL-QUOTE(1)