[Discuss] How to keep '-' in bash args

[pw]

Justin Rebelo wrote:
> pw wrote:
>> Can anyone tell me how to ensure (sh flags ??) that
>> bash doesn't strip '-' from the front end of
>> arguments passed to a shell script?
>>
>> ie: sh script.sh  -myarg
>>
>> #!/bin/bash
>> #script.sh
>> echo "$1"
>>
>>
>> returns:
>> myarg instead of -myarg
> 
> man bash
> search for "--"
> 

On second thought/look the problem does not appear
to be what I thought it was since echoing back the
arguments by value (ie: echo $1 ; echo $2) produces '-'.

The problem seems to be with ${N} where N is the ordinal of the
argument.

ie:

for ARG in `seq 1 256`; do
	echo "${ARG}";
done

Apparently this doesn't work very well with bash
and has become inconsistent with several other shells.

It appears that I should be using 'getopts' instead
of this.

ie:

#!/bin/bash
#test2.sh

while getopts .flag1:flag2:flag3:flagn. OPTION
	case $OPTION in
		flag1)
			echo $OPTARG;
		;;
		flag2)
			echo $OPTARG;
		;;
		flag3)
			echo $OPTARG;
		;;
		flagn)
			echo $OPTARG;
		;;
		?)
			echo "bad arg";
	esac
done


test2.sh -flag1 one -flag2 two -flag3 three -flagn en -flagz

outputs:

one
two
three
en
bad arg

Peter