CODEWITHSUNDEEP
bashloopsarguments
Alexander
Alexander

Reputation: 1378

Looping through arguments in bash

I'm trying to write a code that runs on my arguments list , for example if I have -p a b c d -q g e f c as arguments: when I get -p, I want the loop to run on variables a b c d, until I get -q, and then to do something else, likewise I want it to be in reverse;

this is my code:

#bin/bash
while test -n "$1" -a ${1:0:1} = - ;do
if test x$1=x-q:then
    shift
    while test -n "$1" ; do
        echo $1
        if test x$2=x-p;then 
            break;
        shift
    done
fi
if test x$1=x-p;then 
   echo 'print test'+$1;
   shift
fi
done

but break doesn't seem to work, does anyone know how I can implement this?

Upvotes: 0

Views: 78

Answers (1)

glenn jackman
glenn jackman

Reputation: 246754

Consider first parsing all the arguments, and collection the "-p" args in one array and the "-q" args in another array:

p_args=() 
q_args=()
opt=""

for arg do 
    case $arg in 
        "-p") opt=p ;; 
        "-q") opt=q ;; 
           *) [[ $opt == p ]] && p_args+=("$arg")
              [[ $opt == q ]] && q_args+=("$arg")
              ;; 
    esac
done

# do stuff with "-p" args
declare -p p_args

# do stuff with "-q" args
declare -p q_args

Upvotes: 2

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