ssh command with quotes

Question

I have an odd error that I have been unable to find anything on this. I wanted to change the user comment with the following command.

$ sudo usermod -c "New Comment" user

This will work while logged onto a server but I want to automate it across 20+ servers. Usually I am able to use a list and loop through the servers and run a command but in this case I get a error.

$ for i in `cat servlist` ; do echo $i ; ssh $i sudo usermod -c "New Comment" user ; done 
serv1
Usage: usermod [options] LOGIN
Options:
lists usermod options
serv2
Usage: usermod [options] LOGIN
Options:
lists usermod options
.
.
.

When I run this loop it throws back an error like I am using the command incorrectly but it will run just fine on a single server.

Looking through the ssh man pages I did try -t and -t -t flags but those did not work.

I have successfully used perl -p -i -e within a similar loop to edit files.

Does anyone know a reason I am unable to loop this?

Answer 1

SSH executes the remote command in a shell. It passes a string to the remote shell, not a list of arguments. The arguments that you pass to the ssh commands are concatenated with spaces in between. The arguments to ssh are sudo, usermod, -c, New Comment and user, so the remote shell sees the command

sudo usermod -c New Comment user

usermod parses Comment as the name of the user and user as a spurious extra parameter.

You need to pass the quotes to the remote shell so that the comment is treated as a string. The simplest way is to put the whole remote command in single quotes. If you need a single quote in that command, use '\''.

ssh "$i" 'sudo usermod -c "Jack O'\''Brian" user'

Instead of calling ssh in a loop and ignoring errors, use a tool designed to run commands on multiple servers such as pssh, mussh, clusterssh, etc. See Automatically run commands over SSH on many servers

• Charlie Dalsass: Another good tool is Ansible for running commands on multiple servers.

Answer 2

for i in `cat servlist`;do echo $i;ssh $i 'sudo usermod -c "New Comment" user';done

or

for i in `cat servlist`;do echo $i;ssh $i "sudo usermod -c \"New Comment\" user";done

Answer 3

You can use following convenient wrapper script ssh.sh

EDIT 2023年04月28日. Finally I figure out the perfect solution, fixed the issue mentioned by @user202729, yet not over-programing.

The final ssh wrapper is:

#!/bin/bash
args=(); for v in "$@"; do args+=("$(printf %q "$v")"); done
ssh "${args[@]}"

You can create it by copy&paste run:

cat <<'EOF' > ssh.sh
#!/bin/bash
args=(); for v in "$@"; do args+=("$(printf %q "$v")"); done
ssh "${args[@]}"
EOF
chmod +x ssh.sh

Then you can safely call ssh via the ssh.sh, without worrying about escaping.

./ssh.sh host sudo usermod -c "New Comment" user

A full test:

First create a utility /tmp/show_args.sh which shows all arguments

cat <<'EOF' > /tmp/show_args.sh
#!/bin/bash
for arg in "$@"; do echo "ARG$((++i))=${arg@Q}"; done
EOF
chmod +x /tmp/show_args.sh

The do the full test:

./ssh.sh 127.0.0.1 -n /tmp/show_args.sh "a a" "'b b'" '"c c"' '*' '()' $'line1\nline2' $'001円 a' 'zz '

The output is:

ARG1='a a'
ARG2=''\''b b'\'''
ARG3='"c c"'
ARG4='*'
ARG5='()'
ARG6=$'line1\nline2'
ARG7=$'001円 a'
ARG8='zz '

You can see that all arguments are same as the input. Note

''\''b b'\'''

just means literal

'b b'

• user202729: This will give wrong result if the output of %q contains two consecutive spaces e.g. it happens for echo $'\x01 12 3' in my bash version
• osexp2000: oh sorry to here that. Maybe it is because of bash version. Could you test it with following command? I have tried it, no problem. ./ssh.sh host echo $'\x01 12 3' | od -txCc, then check the binary output, it should be 01 20 31 32 20 33 0a.
• dave_thompson_085: But echo $'\x01 12 3' executed locally is 01 20 20 31 32 20 20 33 0a, and your theory is it will be the same remotely, but it's not, because command substitution $( ) breaks at whitespace -- even if it occurs between quotemarks, unlike shell input. This applies to all bash versions (although very old versions don't have %q and never reach the point of error). More dramatically, try ssh.sh host echo $'\x07 * * * IMPORTANT * * *'.
• osexp2000: @dave_thompson_085 nice catch, thanks for telling me this, I will find a workaround.
• user202729: No need, overcomplicated. Just change ssh $(escape "$@") to ssh "$(escape "$@")" in the first script (didn't test but should work.