CODEWITHSUNDEEP
linuxshellunixsh
thundium
thundium

Reputation: 1053

how to loop argument in sh shell

I know "for loop" can be used to go through all arguments. But for my code, I would prefer

i = 1
if [ ${$i} == xx] ; then
 xxx
fi

((iArg++))

if [ ${$i} == xx] ; then
 xxx
fi

Oberviously, ${$i} is not working. How shall I make it correct?

I used to write csh by using 

$argv[$i]

Got to use sh shell now. Thanks for help

Upvotes: 2

Views: 231

Answers (1)

John Kugelman
John Kugelman

Reputation: 361564

Because of this very difficulty, argument parsing is typically done by using $1 and shift. A typical loop might look like:

while [ $# -gt 0 ]; do
    case $1 in
        -a) echo "-a";    shift 1;;   # option with no argument
        -b) echo "-b $2"; shift 2;;   # option which takes argument, use $2
        -c) echo "-c";    shift 1;;
    esac
done

That said, the way to do indirect variable names is with !, as in:

${!i}

Upvotes: 5

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