bash - Use Variable with multiple options in Case statement -


is possible following (and how ?)(,resp. why not ?):

match="--opt1 | --opt2" while true ;     case $1 in         $match)              echo "option $2" found;              shift 2;;         *)              unknown option; exit 1;     esac done

for reason dont understand doesnt work. however, having 1 alternative match="--opt1" fine.

edit 1: possible solution

instead of going case statement 1 check if given option occurs in string of multiple allowed options, instance using grep , if. full dynamically 1 consider follwoing solution, might combined or embedded within case statement:

while true ; if [ -n "$(echo $matches|grep -- $1)" ];     echo "found option $1 value $2"     shift 2 fi done

the pipe character, when embedded in parameter value, treated literally, not syntax. you'll have use multiple strings:

while true;     case $1 in        $match1 | $match2 )        # etc

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