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

[pw]

noel at natnix.com wrote:
> if $1 is set to "-myarg", then 
> 
> 	echo "$1"
> 
> expands to 
> 
> 	echo -myarg
> 
> and echo will attempt to interpret "-myarg" as an option, instead of
> printing it.
> 
> This problem is quite fun when you try to remove a file named "-".
> 
> Most commands accept "--" to signal the end of options, so this would
> work for you:
> 
> 	echo -- "$1"
> 
> But hold on!  echo does not support a "--" argument!  (I guess it does pay to
> try your own advice before shooting your mouth off.) "echo -- -a" prints
> "-- -a".  but "echo -a" prints "".
> 
> You could do a hack like this:
> 
> 	echo "" -a
> 
> which prints " -a", but that leading space is annoying.
> 
> This might be a little more bullet-proof:
> 
> 	printf '%s\n' "$1"
> 
> --Noel

This works but I have to put all my args into a single string:

#!/bin/bash
#test3.sh

for ARG in $1;do
         FLAG=`printf '%s\n' "$ARG" |sed -e "s/\ //g" | \
	awk '{A=index($0,"-");if(A==1){print 1}else{print 0 " " $0}}'`;

	if [ "$FLAG" = "1" ]; then
                 echo "$ARG flag";
         else
                 echo "$ARG not flag";
         fi
done






run:

sh test3.sh "-flag1 value1 -flag2 value2 -flagn valuen"


output:

-flag1 flag
value1 not flag
-flag2 flag
value2 not flag
-flagn flag
valuen not flag


Peter