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

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 2023/04/28. 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    a' 'zz   '

The output is:

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

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

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

just means literal

'b b'