bash typeset issue execute function on remote server

Question

I am trying to use typeset command to execute the function remotely. but unfornately it exit with error.

i have two functions one is for user creation and one that will use to run function remotely

##./create_user -u test1 -g staff -c "test test"  -h host1
uadd() {
  if grep -E "$username" /etc/passwd >/dev/null; then
    echo "$username user already exists in the server $host"
    exit 1
  else

    if grep -E "$groupname" /etc/group >/dev/null; then
      /usr/sbin/useradd -G staff,"$groupname" -c "$comment" "$username"
      echo "$username user created with staff and $groupname groups!"
      exit 1
    else
      /usr/sbin/useradd -G staff -c "$comment" "$username"
      echo "$username user created with staff group only!"
      exit 1
    fi
  fi

}

remoteuadd() {

  #ssh sysadm@$host "$(typeset -f useradd); useradd"
  ssh root@"$host" -p 5829 "$(declare -f uadd);uadd"

  #ssh root@"$host" -p 5829 "$(printf "%q " bash -c "$(declare -f uadd); uadd")"

  #typeset -f uadd | ssh -v  root@$host -p 5829  `$(cat); `

  #ssh root@"$host" -p 5829 bash -s <<EOF
  #$(declare -f uadd)
  #uadd
  #EOF

}

its working but result looks like this

+ remoteuadd
+++ typeset -f uadd
++ uadd '()' '{' egrep '"$username"' /etc/passwd '>' '/dev/null;' if '[' '$?' == 0 '];' then echo '"$username' user already exists in the server '$hostname";' exit '1;' else egrep '$groupname' /etc/group '>' '/dev/null;' if '[' '$?' == 0 '];' then /usr/sbin/useradd -G 'staff,$groupname' -c '"$comment"' '$username;' echo '"$username' user created with '$groupname' 'groups!";' exit '1;' else /usr/sbin/useradd -G staff -c '"$comment"' '$username;' echo '"$username' user created with staff group 'only!";' exit '1;' 'fi;' fi '}'
++ egrep test1 /etc/passwd
++ '[' 0 == 0 ']'
++ echo 'test1 user already exists in the server '
++ exit 1
+ ssh root@localhost -p 5829 test1 user already exists in the server
bash: test1: command not found
+ exit 1

as you can see script try to create user if not exits and so it works and found that user already exits but still script didn't exit it catch the result value and run again and ultimately script break with error.

Any suggestion highly appreciated.

Regards, Mir

Answer 1

to execute the function remotely

Yet, you are using ` backticks which execute stuff locally. You are exeucuting $(typeset -f uadd);uadd and then executing ssh with the result of the function.

The most understandable is a redirection:

ssh root@"$host" -p 5829 bash -s <<EOF
$(declare -f uadd)
uadd
EOF

Note that ssh kind-of-evals the argument. You can also double-escape for ssh:

ssh root@"$host" -p 5829 "$(printf "%q " bash -c "$(declare -f uadd); uarr")"

Notes:

• do not use backticks. Use $(..) instead.
• check your script with shellcheck.
• Remember to quote parameter expansions
• the result you posted does not match the code you posted at all, it's unrelated
• typeset is something for portability with ksh shell. In Bash, use declare.
• https://mywiki.wooledge.org/BashFAQ/096
• You might be interested in https://stackoverflow.com/a/68359474/9072753 and some my script https://gitlab.com/Kamcuk/kamilscripts/-/blob/master/etc/bash.d/sshrfunc.sh