Shell Programming and Scripting

I ran the following program. I don't understand some of it.

#!/bin/sh
case $1 in
--test|-t)
echo "you used the --test option"
exit 0
;;
--help|-h)
echo "Usage:"
echo " myprog.sh [--test|--help|--version]"
exit 0
;;
--version|-v)
echo "myprog.sh version 0.0.1"
exit 0
;;
-*)
echo "No such option $1"
echo "Usage:"
echo " myprog.sh [--test|--help|--version]"
exit 1
;;
esac

echo "You typed \"$1\" on the command-line"

-----------------------------------------------------------------------------

Do I need a program called '' myprog.sh '' to run this program?

I got the following message when running it.

You typed "" on the command-line.

The name of the program is '' temp25 '' .


[nissanka@c83-250-111-20 ~]$ chmod 755 temp25
[nissanka@c83-250-111-20 ~]$ ./temp25
You typed "" on the command-line

--------------------------------------------------
I have a file called '' Cisco ''

[nissanka@c83-250-111-20 ~]$ ./temp25 Cisco
You typed "Cisco" on the command-line
[nissanka@c83-250-111-20 ~]$


Could you please tell me how to run this program and how it works?

You are running the program alright. You do not need anything else to run this program.
What this program does is check for what options were passed in to it. It does this by checking for $1 (the first option/arg passed) in the case statement. If you send in any of the options checked for in the case statement, it will execute the commands that are below each of the cases upto the ';;' i.e.
If you run the program as "./temp25 --test", the commands executed will be 'echo "you used the --test option"
exit 0',
and so on.

If you send in as arguments any thing that is not present in the case statement (without a leading '-'), it will just execute the last statement in the script:

Code:

echo "You typed \"$1\" on the command-line"