基于第一个值的不同选项的 bash 完成

Question

I'm currently trying to write a bash_completion script for one of our tools. Was looking at the apt-get and chkconfig scripts for some help. Basically what I want to do is get a different option selection based on the first value. There can be more than one option to a value.

command <value1> <--option1> <--option2> ...
command <value2> <--option3> <--option4> ...

Looking at the apt-get script, it will return the same list of options for any first value. So this is not what I need.

Here is what I got so far:

 _supdeploy()
{
    local cur prev opts cword
    _init_completion || return

    COMPREPLY=()
    cur="${COMP_WORDS[COMP_CWORD]}"
    prev="${COMP_WORDS[COMP_CWORD-1]}"
    opts="deploy destroy supported list fulllist status fullstatus getip shutdown powerOff powerOn"

    case "${prev}" in
        deploy)
           if [[ "$cur" == -* ]]; then
               if [[ $cword -eq 2 || $cword -eq 3 || $cword -eq 4 ]]; then
                   COMPREPLY=( $( compgen -W '--hostname --os --version' -- "$cur" ) )
               fi
           fi
           return 0
           ;;
        destroy)
            if [[ "$cur" == -* ]]; then
               COMPREPLY=( $( compgen -W '--name --silent' -- "$cur" ) )
            fi
            return 0
            ;;
        *)
            COMPREPLY=( $(compgen -W "${opts}" -- ${cur}) )
            return 0
            ;;
    esac

} &&
complete -F _supdeploy supdeploy

I get a different selection of options for both deploy and destroy. But I can only use one -- option. When I try to use -- again, nothing happens, and without the -- I get the list from opts.

It is probably something easy, but I can't find the error here at the moment. I also have it tried without the $cword before, same result

Answer 1

Instead of prev , you just want to look at the value of the first argument each time:

case ${COMP_WORDS[1]} of

It gets a little tricker if you allow "general" options to precede the subcommand, but in general you want to look at the first non-option argument, not necessarily the previous argument.