如何为路径上的所有可执行文件编写bash完成脚本？

Question

I've had this use case come up for a couple of different scripts I've written or modified. Essentially, I want bash completion for option '-x' to complete executables on the PATH. This is sort of two questions wrapped in one.

So far I've had troubles because bash doesn't easily distinguish between aliases, builtins, functions, etc and executable files on the PATH. The _commands wrapper function in /usr/share/bash-completion/bash_completion completes on all of the above but I have no use for working with aliases, builtins, functions, etc and only want to complete on the commands that happen to be executables on the PATH.

So for example... If I enter scriptname -x bas[TAB] , it should complete with base64, bash, basename, bashbug.

This is what my completion script looks like now:

_have pygsparkle && {

_pygsparkle(){
    local cur prev

    COMPREPLY=()
    cur=${COMP_WORDS[COMP_CWORD]}
    prev=${COMP_WORDS[COMP_CWORD-1]}

    case $prev in
    -x|--executable)
        # _command
        executables=$({ compgen -c; compgen -abkA function; } | sort | uniq -u)
        COMPREPLY=( $( compgen -W "$executables" -- "$cur" ) )
        return 0
        ;;
    esac

    if [[ $cur = -* ]]; then
        COMPREPLY=( $( compgen -W '--executable -h --help -x' -- "$cur" ) )
    fi

}

complete -F _pygsparkle pygsparkle

}

It seems to work as expected but { compgen -c; compgen -abkA function; } | sort | uniq -u { compgen -c; compgen -abkA function; } | sort | uniq -u { compgen -c; compgen -abkA function; } | sort | uniq -u is a pretty dirty hack. In zsh you can get a sorted list of executables on PATH running print -rl -- ${(ko)commands} . So it appears I'm missing at least 60+ execs, likely because uniq -u is dumping execs with that same name as aliases or functions.

Is there a better way to do this? Either a better command for getting all executables on PATH or a pre-existing completion function that will serve the same ends?

Update: Ok so the following function executes in under 1/6 sec and looks like the best option. Unless there are any other suggestions I'll probably just close the question.

_executables(){
    while read -d $'\0' path; do
        echo "${path##*/}"
    done < <(echo -n "$PATH" | xargs -d: -n1 -I% -- find -L '%' -maxdepth 1 -mindepth 1 -type f -executable -print0 2>/dev/null) | sort -u
}

Answer 1

I have a run-in-background script . To get autocomplete for it I use bash's builtin complete :

> complete -c <script-name>
> <script-name> ech[TAB]
> <script-name> echo 

From the docs :

 command Command names. May also be specified as -c. 

---

To deal with non-executables, my first thought was to filter with which or type -P but that's terribly slow. A better way is to only check the unknowns. Use @Six's solution or some other to find items in compgen -c but not compgen -abkA function . For everything in both lists, check with type -P . For example:

function _executables {
    local exclude=$(compgen -abkA function | sort)
    local executables=$(
        comm -23 <(compgen -c) <(echo $exclude)
        type -tP $( comm -12 <(compgen -c) <(echo $exclude) )
    )
    COMPREPLY=( $(compgen -W "$executables" -- ${COMP_WORDS[COMP_CWORD]}) )
}
complete -F _executables <script-name>

I've tried a few times but introducing -F to complete just seems to be a slow way to do thing. Not to mention now having to handle absolute/relative path completion to executables, the horribly fiddly complete-the-directory-but-don't-add-a-space bit and correctly handling spaces in paths.

Answer 2

看起来你正在寻找compgen -c

Answer 3

There doesn't seem to be a simple answer to the question of how to list all the executables available on a user's PATH. Alas, I have searched far and wide for an answer.

compgen may at first seem like the right direction but it lacks an option to show only executables

compgen -c will show all commands (this includes aliases, builtins, executables, functions, and keywords)

compgen -abkA function will show all commands except executables

So an approximation of the executables available can be surmised by 'diffing' the two, ie { compgen -c; compgen -abkA function; } | sort | uniq -u { compgen -c; compgen -abkA function; } | sort | uniq -u { compgen -c; compgen -abkA function; } | sort | uniq -u , but that clearly has some issues.

NOTE: To confirm that compgen -c will not work, all you have to do is scan through the results and identify many non-executable entries. You can also try diff -u <(compgen -A function -abck | sort -u) <(compgen -c | sort -u) and see that the commands are equivalent (once duplicates have been removed of course).

So it seems the best option is to scan through every directory on the path.

In summary, the following is the best option on a modern system. It has a speedy runtime (~0.05 seconds) comparable to compgen -c and suitable for completions.

executables(){
    echo -n "$PATH" | xargs -d: -I{} -r -- find -L {} -maxdepth 1 -mindepth 1 -type f -executable -printf '%P\n' 2>/dev/null | sort -u
}

If you are using ancient versions of find/xargs (ie busybox or BSD) and/or using mksh (Android's default shell) which doesn't support process substitution, you will want the following. Not so speedy (~3.5 seconds).

executables(){
    find ${PATH//:/ } -follow -maxdepth 1 -type f -exec test -x '{}' \; -print0 2>/dev/null \
    | while read -d $'\0' path; do
        echo "${path##*/}"
    done \
    | sort -u
}

If you are using the aforementioned setup and have spaces in your PATH for some reason, then this should work.

executables(){
    echo "$PATH" \
    | tr ':' '\n' \
    | while IFS= read path; do
        find "$path" -follow -maxdepth 1 -type f -exec test -x '{}' \; -print0 2>/dev/null
    done \
    | while read -d $'\0' path; do
        echo "${path##*/}"
    done \
    | sort -u
}