Scripting question

jim.thornton · 03-21-2008, 11:05 PM

Is it possible when writing a bash script to have a case statement that can handle a range?? Or would I have to use multiple if statements.

For example, I have to write a script that will count the number of arguments provided. I can do it with if statements but I would like to do it with a case statment and have three conditions:

1. > 4 ---> If more than 4 arguments are provided do...
2. < 4 ---> If less than 4 arguments are provided do...
3. = 4 ---> If 4 arguments are provided do...

Is this possible with case??

colucix · 03-22-2008, 04:09 AM

You can do something like:

Code:

case $# in

   [0-3])   echo "less than 4" ;;

   4)   echo  "equal to 4" ;;

   *)   echo  "more than 4" ;;

esac

Here the [0-3] is not really for numerical comparison, but it is a range of digits. It would be a little more complicated if the range was between larger numbers (e.g. 23-47) but in your case it should do the trick.

jim.thornton · 03-22-2008, 07:56 AM

Quote:

Originally Posted by colucix

You can do something like:

Code:

case $# in

   [0-3])   echo "less than 4" ;;

   4)   echo  "equal to 4" ;;

   *)   echo  "more than 4" ;;

esac

Here the [0-3] is not really for numerical comparison, but it is a range of digits. It would be a little more complicated if the range was between larger numbers (e.g. 23-47) but in your case it should do the trick.

Awesome... Thank you.

From your answer it seems as though there is no way to test if there is more than x or less than x. Am I correct. I would always have to improvise like that?

colucix · 03-22-2008, 09:13 AM

Quote:

Originally Posted by jim.thornton

From your answer it seems as though there is no way to test if there is more than x or less than x. Am I correct. I would always have to improvise like that?

Correct. The case statement compares the value of the variable with those provided inside the construct. It performs a "string comparison" and as far as I know there is no way to use comparison operators.

matthewg42 · 03-22-2008, 09:25 AM

Case uses a pattern to determine which section of the case statement to execute. For single digits, it's simple to use patterns like those above which will allow you to effectively select by number range, but for more complex numbers it is not easy. Consider numbers more than 10... then the patterns become much harder to make, at least in a way which is not very long winded.

But you have if...elif... fi which can be used very well for this:

Code:

if [ $number -ge 0 ] && [ $number -lt 10 ]; then
    echo "$number is in range 0..9"
elif [ $number -ge 10 ] && [ $number -lt 20 ]; then
    echo "$number is in range 10..19"
elif [ $number -ge 20 ] && [ $number -lt 30 ]; then
    echo "$number is in range 20..29"
fi

jim.thornton · 03-23-2008, 05:58 PM

Thanks...

How do I do an OR (||) when scripting for bash? I tried using || and it didn't work

matthewg42 · 03-23-2008, 06:08 PM

You can use || as or. You have to make sure you put spaces in the right places. What did you try?

jim.thornton · 03-23-2008, 07:24 PM

Quote:

Originally Posted by matthewg42

You can use || as or. You have to make sure you put spaces in the right places. What did you try?

I want it to take -i or -I as the argument and this is what I have:

Code:

if [ "$1" != "-i" || "$1" != "-I" ]

The this is the error I'm getting:

./test.sh: line 50: [: missing `]'
./test.sh: line 50: -i: command not found

matthewg42 · 03-24-2008, 04:49 AM

The syntax would be like this:

Code:

if [ "$1" != "-i" ] || [ "$1" != "-I" ]

If you're parsing command line options, you might want to take a look at the bash internal getopt (edit: and getopts) which can be used to validate options for you.

For example, if you have a script which can take the options -a -b and -C, where -b takes an argument to the option, you might do it like this:

Code:

#!/bin/bash

while getopts "ab:C" option; do
        case $option in
        a|C)
                echo "we have option: $option (no argument for this option)"
                ;;
        b)
                echo "we have option: $option with argument: $OPTARG"
                ;;
        *)
                echo "ERROR - unknown option" >&2
                exit 1
                ;;
        esac
done

# remove options from parameter list, leaving just remaining parameters
shift $(($OPTIND - 1))

for p in "$@"; do
        echo "post-option parameter: $p"
done

An example invocation and output:

Code:

% ./test.sh -a -b "my option argument" -C now the rest of the parameters
we have option: a (no argument for this option)
we have option: b with argument: my option argument
we have option: C (no argument for this option)
post-option parameter: now
post-option parameter: the
post-option parameter: rest
post-option parameter: of
post-option parameter: the
post-option parameter: parameters

jim.thornton · 03-24-2008, 08:03 AM

Wow... That was such a great response.

Thank you.

smehra24 · 03-24-2008, 08:12 AM

case $# in
[0-3]) echo 'arguments less than 4';;
4) echo 'arguments equal to 4';;
*) echo 'More than 4'
esac