为什么总是执行 bash 脚本默认 case 语句？

Question

I wrote a bash script that reads script arguments and pass them to parsearg function:

#!/usr/bin/env bash
function main() {
    parseargs2 "$@"
}

function parseargs2() {
    MAINCOMMAND=$1
    shift
    while [ $# -gt 0 ]; do
        case $1 in
            -s|--service) SERVICE_NAME="$2" ;;
            -r|--registry) REGISTRY="$2" ;;
            -h|--help) HELP=true ;;
            *) echo "help" && exit 1;;
        esac
        shift
    done
    echo "SERVICE_NAME: $SERVICE_NAME"
    echo "REGISTRY: $REGISTRY"
    echo "HELP: $HELP"
    echo "cmd: $MAINCOMMAND"
}

main "$@"

now when I run my script it always executes help command and then exits, I don't know why it will be ok when I remove *) case

./example.sh build --service api --registry dockerhub

EDIT: thanks to @chepner comment I found the problem I solved this by adding shift 2 at end of while loop

Answer 1

Doing your own option parsing is a little fraught, but the idea should be to do an extra shift whenever you have an argument:

while [ $# -gt 0 ]; do
    case $1 in
        -s|--service) SERVICE_NAME="$2" ; shift;;
        -r|--registry) REGISTRY="$2" ; shift;;
        -h|--help) HELP=true ;;
        *) echo "help" && exit 1;;
    esac
    shift
done

You may or may not want to support a style where the single-letter version of an option has its argument jammed up against it, which works in many stock programs:

while [ $# -gt 0 ]; do
    case $1 in
        -s|--service) SERVICE_NAME="$2" ; shift;;
        -s*) SERVICE_NAME=${1#-s};;
        -r|--registry) REGISTRY="$2" ; shift;;
        -r*) REGISTRY=${1#-r};;
        -h|--help) HELP=true ;;
        *) echo "help" && exit 1;;
    esac
    shift
done

Also, there's no reason to name your variables in all-caps. All caps is usually a sign that a variable is being export ed into the environment so some other program you're going to run can see it; otherwise, just use lowercase. Additionally, you don't need quotation marks on the right-hand side of an assignment, and if you're really using bash (rather than some other POSIX type shell), you likely want to use (( ... )) for numeric comparisons.

help=   # don't assume variables aren't set on entry
maincommand=$1
shift
while (( $# )); do
    case "$1" in
        -s|--service) service_name=$2 ; shift;;
        -s*) service_name=${1#-s};;
        -r|--registry) registry=$2 ; shift;;
        -r*) registry=${1#-r};;
        -h|--help) help=true ;;
        *) echo "help" && exit 1;;
    esac
    shift
done
for var in service_name registry help maincommand; do
   printf '%s: "%s"\n' "$var" "${!var}"
done

FWIW, true is just a string to the shell with no special significance; the shell doesn't have Boolean values beyond "command succeeded/failed" (which is just zero/nonzero on the exit code) or Boolean operators other than the ones that operate on commands (eg command1 && run this if command1 succeeded ). So when flags are stored in shell variables the Booleanness is usually represented either by empty vs. nonempty string (and true is a fine value to use for the nonempty case) or 0 vs. 1 (or 0 vs nonzero) number.