Non-interactive SSH password authentication

Vincent Bernat

SSH offers several forms of authentication, such as passwords and public keys. The latter are considered more secure. However, password authentication remains prevalent, particularly with network equipments.1

A classic solution to avoid typing a password for each connection is sshpass, or its more correct variant passh. Here is a wrapper for Zsh, getting the password from pass, a simple password manager:2

pssh() {
  passh -p <(pass show network/ssh/password | head -1) ssh "$@"
}
compdef pssh=ssh

This approach is a bit brittle as it requires to parse the output of the ssh command to look for a password prompt. Moreover, if no password is required, the password manager is still invoked. Since OpenSSH 8.4, we can use SSH_ASKPASS and SSH_ASKPASS_REQUIRE instead:

ssh() {
  set -o localoptions -o localtraps
  local passname=network/ssh/password
  local helper=$(mktemp)
  trap "command rm -f $helper" EXIT INT
  > $helper <<EOF
#!$SHELL
pass show $passname | head -1
EOF
  chmod u+x $helper
  SSH_ASKPASS=$helper SSH_ASKPASS_REQUIRE=force command ssh "$@"
}

If the password is incorrect, we can display a prompt on the second tentative:

ssh() {
  set -o localoptions -o localtraps
  local passname=network/ssh/password
  local helper=$(mktemp)
  trap "command rm -f $helper" EXIT INT
  > $helper <<EOF
#!$SHELL
if [ -k $helper ]; then
  {
    oldtty=\$(stty -g)
    trap 'stty \$oldtty < /dev/tty 2> /dev/null' EXIT INT TERM HUP
    stty -echo
    print "\rpassword: "
    read password
    printf "\n"
  } > /dev/tty < /dev/tty
  printf "%s" "\$password"
else
  pass show $passname | head -1
  chmod +t $helper
fi
EOF
  chmod u+x $helper
  SSH_ASKPASS=$helper SSH_ASKPASS_REQUIRE=force command ssh "$@"
}

A possible improvement is to use a different password entry depending on the remote host:3

ssh() {
  # Grab login information
  local -A details
  details=(${=${(M)${:-"${(@f)$(command ssh -G "$@" 2>/dev/null)}"}:#(host|hostname|user) *}})
  local remote=${details[host]:-details[hostname]}
  local login=${details[user]}@${remote}

  # Get password name
  local passname
  case "$login" in
    admin@*.example.net)  passname=company1/ssh/admin ;;
    bernat@*.example.net) passname=company1/ssh/bernat ;;
    backup@*.example.net) passname=company1/ssh/backup ;;
  esac

  # No password name? Just use regular SSH
  [[ -z $passname ]] && {
    command ssh "$@"
    return $?
  }

  # Invoke SSH with the helper for SSH_ASKPASS
  # […]
}

It is also possible to make scp invoke our custom ssh function:

scp() {
  set -o localoptions -o localtraps
  local helper=$(mktemp)
  trap "command rm -f $helper" EXIT INT
  > $helper <<EOF 
#!$SHELL
source ${(%):-%x}
ssh "\$@"
EOF
  command scp -S $helper "$@"
}

For the complete code, have a look at my zshrc. As an alternative, you can put the ssh() function body into its own script file and replace command ssh with /usr/bin/ssh to avoid an unwanted recursive call. In this case, the scp() function is not needed anymore.

Update (2023-12)

This post was heavily discussed on Hacker News.

  1. First, some vendors make it difficult to associate an SSH key with a user. Then, many vendors do not support certificate-based authentication, making it difficult to scale. Finally, interactions between public-key authentication and finer-grained authorization methods like TACACS+ and Radius are still uncharted territory. ↩︎

  2. The clear-text password never appears on the command line, in the environment, or on the disk, making it difficult for a third party without elevated privileges to capture it. On Linux, Zsh provides the password through a file descriptor. ↩︎

  3. To decipher the fourth line, you may get help from print -l and the zshexpn(1) manual page. details is an associative array defined from an array alternating keys and values. ↩︎