Bash completion function on various OSs: problem with @ while completing hosts

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I'm writing a function that mimics the ssh command host completion; the main difference being that it uses the hosts of the xyz.com domain found in ~/.ssh/known_hosts for the completion.

Here is the function:

_xyzssh() {
    COMPREPLY=()

    local arg="${COMP_WORDS[COMP_CWORD]}"

    [[ $arg != -* ]] || return 0

    [ -f ~/.ssh/known_hosts ] || return 0

    local -a known_hosts
    read -d '' -a known_hosts < <(awk '{
        n = split($1,arr,",");
        for (i = 1; i <= n; i++)
            if (arr[i] ~ /\.xyz\.com$/)
                print arr[i] 
    }' ~/.ssh/known_hosts)

    if [[ $arg == *@* ]]
    then
        known_hosts=( ${known_hosts[@]/#/${arg%%@*}@} )
    fi

    COMPREPLY=( $(compgen -W "${known_hosts[*]}" -- $arg) )

    return 0
}
complete -F _xyzssh xyzssh

This code works fine in Linux (bash 4.2) but it has a weird behavior in macOS (bash 3.2). For example, when I type:

xyzssh -X user@server Tab

  • On Linux it completes with: xyzssh -X user@server.xyz.com

  • On macOS it "doubles" the username: xyzssh -X useruser@server.xyz.com

I'm guessing that there is something fishy with the @ because when I type:

xyzssh -X user\@server Tab

It completes correctly on both Linux and macOS.

Is there a way to fix this behavior on macOS? fixed: see accepted answer below

Update

Just found out about COMP_WORDBREAKS, which contains @ in macOS while in Linux it doesn't.

I can fix the issue when I set COMP_WORDBREAKS="${COMP_WORDBREAKS//@}" in the shell, but making the change resilient is not that easy in macOS (exporting this setting in my .bash_profile or .bashrc doesn't seem to work)...

Update

I got other issues, with Solaris (bash 4.4) this time, in which ${COMP_WORDS[COMP_CWORD]} doesn't even contain the user@ part when the completion function is called... I' m still investigating how to fix it. fixed: see accepted answer below

It's my first time writing a completion function so I might be missing something obvious.

solaris macos linux bash
1 Answers

Taking the path of answering my own question

COMPREPLY needs to be populated differently depending on the presence of @ in COMP_WORDBREAKS.

For example, when the user type ssh user@server Tab:

  • if @ is not in COMP_WORDBREAKS then COMPREPLY should look like:
COMPREPLY=( "user@server.xyz.com" "user@server-bis.xyz.com" )
  • if @ is in COMP_WORDBREAKS then COMPREPLY should look like:
COMPREPLY=( "@server.xyz.com" "@server-bis.xyz.com" )

A solution for making the completion function compatible with both cases is to add the right prefix after finding the matching hosts list:

Update: fixed Solaris bash 4.4 issue

_xyzssh() {
    COMPREPLY=()

    local curr="${COMP_WORDS[COMP_CWORD]}"
    local prev="${COMP_WORDS[COMP_CWORD-1]}"

    [[ $curr == -* ]] && return 0

    [ -f ~/.ssh/known_hosts ] || return 0

    local -a known_hosts
    IFS=$'\n' read -r -d '' -a known_hosts < <(awk '{
      n = split($1,arr,",");
        for (i = 1; i <= n; i++)
          if (arr[i] ~ /\.xyz\.com$/)
            print arr[i]
    }' ~/.ssh/known_hosts)

    COMPREPLY=( $(compgen -W "${known_hosts[*]}" -- "${curr#*@}") )

    if [[ $curr == *@* || $prev == @ ]]
    then
        if [[ $COMP_WORDBREAKS == *@* ]]
        then
            COMPREPLY=( "${COMPREPLY[@]/#/@}" )
        else
            COMPREPLY=( "${COMPREPLY[@]/#/${curr%%@*}@}" )
        fi
    fi

    return 0
}
complete -F _xyzssh xyzssh
Debug information

Here is what I get in COMP_WORDS when I type xyzssh -X user@server Tab on the different OSs:

  • redhat 7 (bash 4.2, @ not present in COMP_WORDBREAKS)
COMP_WORDS=( "xyzssh" "-X" "user@server" )
  • macOS (bash 3.2, @ present in COMP_WORDBREAKS)
COMP_WORDS=( "xyzssh" "-X" "user@server" )
  • solaris 11.4 (bash 4.4, @ present in COMP_WORDBREAKS)
COMP_WORDS=( "xyzssh" "-X" "user" "@" "server" )
user_33877160
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