bash: How do I identify the filename argument among arguments passed from the command line?

Question

I know if I have a command ./run -c -p '$' -s 10 file.txt
I can write bash script like this

while read line; do 
   # ...
done < $6

But what if the command may or may not have one/some options, maybe look like this

./run -p '$' -s 10 file.txt

or this

./run '$' -s 10 file.txt

Then how can I get the file name in the script?

Answer 1

If the file name always comes at the end of the list of arguments, you can use "${@: -1}" to select the last argument. Taken from: https://stackoverflow.com/a/1854031/3565972

Eg.:

while read line; do 
   ...
done < "${@: -1}"

Answer 2

Use getopts to process the options (that also allows them to be in arbitrary order):

#!/usr/bin/env bash

while getopts cp:s: option; do
    case $option in
    c)
        echo "-c used"
        ;;
    p)
        echo "-p used with argument $OPTARG"
        ;;
    s)
        echo "-s used with argument $OPTARG"
        ;;
    *)
        echo "unknown option used"
        exit 1
        ;;
    esac
done

shift $(( OPTIND - 1 ));

echo "Arguments left after options processing: $@"

Now if you run this:

$ ./test.sh -c -p '$' -s 10 file.txt
-c used
-p used with argument $
-s used with argument 10
Arguments left after options processing: file.txt

Answer 3

If you want to get the filename in your script ./run. Then you can just use $# to get the total of the parameters. And so you can write as below:

eval filename=\$$#

If you just want to get the filename from a line like this: ./run ** ** ** ** file.txt

You can just use awk as below:

line="./run ** ** ** ** file.txt"
echo $line | awk '{print $NF}'

Good Luck!