Shell Programming and Scripting

Hello. I'm having trouble figuring out how to write this script. I'm supposed to have my script start up with specific arguments and if there are missing arguments or invalid arguments, return an error associated to its errorlevel number.
For example, here are a few errorlevel numbers with their event information.

16 Third argument present, but not what expected (“-o”)
17 Fourth argument missing on command line
18 Too many arguments provided on command line (>4)

First off, i'm supposed to execute my shell script with these specific arguments :

Code:

Testscript.sh -i ./input.txt -o ./output.txt

I've added the following to my script to make sure the script is started with those arguments :

Code:

$# = 4

$0 = ./testscript.sh

$1 = "-i"

$2 = "./input.txt"

$3 = "-o"

$4 = "./output.txt"

$@ = " -i ./input.txt -o ./output.txt"

The part i can't seem to figure out is how to get my script to return an error and an errorlevel number. I was thinking maybe using IF statements in conjungtion with the $? variable but i'm not totally sure. Help would be much appreciated.

Thanks.

Code:

if [ $# -lt 4 ]
then
    echo 'Forth argument missing on command line'
    exit 17
fi
 
if [ $# -gt 4 ]
then
    echo 'Too many arguments provided on command line (>4)'
    exit 18
fi
 
if [ "$3" != "-0" ]
then
    echo 'Third argument present, but not what expected ("-o")'
    exit 16
fi

Wow thank you very much. This makes much more sense now. Heh, simple answer but i'm learning.

When printing error messages, you usually want to redirect them to stderr instead of using stdout.

Code:

echo error message 1>&2

Regards,
Alister

Thanks Alister, I originally had the echo statements as comments as the spec didn't seem to call for any error output.

However I think the above is much more usefull, self-documenting and educational for a beginner scripter than:

Code:

[ $# -lt 4 ] && exit 17
[ $# -gt 4 ] && exit 18
[ "$3" != "-0" ] && exit 16