Why is grep not working?

Question

bash

I'm trying to find options among script parameters.

while [ -n "$1" ]
do
arg=`echo $1 | grep ^-[A-Za-z]+`
echo $arg
if [ "$arg" != "" ]
then 
echo "опция"
fi
shift
done

Why is grep not finding options?

Answer 1

Because the pattern must be in quotation marks. And in grep, you must explicitly indicate that the pattern is a regular expression with the -E option
Do
grep - E 'ваш паттерн'

Answer 2

As noted in the comments, it is better to work with options differently. For example using case like this . I encourage you to check out the other examples as well.