LinuxQuestions.org - Switch statement for bash script

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Switch statement for bash script

Hello, i am currently trying out the switch statement using bash script.

CODE:
showmenu () {
echo "1. Number1"
echo "2. Number2"
echo "3. Number3"
echo "4. All"
echo "5. Quit"
}

while true
do
showmenu
read choice
echo "Enter a choice:"
case "$choice" in
"1")
echo "Number One"
;;
"2")
echo "Number Two"
;;
"3")
echo "Number Three"
;;
"4")
echo "Number One, Two, Three"
;;
"5")
echo "Program Exited"
exit 0
;;
*)
echo "Please enter number ONLY ranging from 1-5!"
;;
esac
done

OUTPUT:
1. Number1
2. Number2
3. Number3
4. All
5. Quit
Enter a choice:

So, when the code is run, a menu with option 1-5 will be shown, then the user will be asked to enter a choice and finally an output is shown. But it is possible if the user want to enter multiple choices. For example, user enter choice "1" and "3", so the output will be "Number One" and "Number Three". Any idea?

put "|" . eg 1|3) echo "1 or 3" ;;

Just something to get you started.

Code:

#! /bin/bash

showmenu ()

{

    typeset ii

    typeset -i jj=1

    typeset -i kk

    typeset -i valid=0        # valid=1 if input is good



    while (( ! valid ))

    do

        for ii in "${options[@]}"

        do

            echo "$jj) $ii"

            let jj++

        done

        read -e -p 'Select a list of actions : ' -a answer

        jj=0

        valid=1

        for kk in "${answer[@]}"

        do

            if (( kk < 1 || kk > "${#options[@]}" ))

            then

                echo "Error Item $jj is out of bounds" 1>&2

                valid=0

                break

            fi

            let jj++

        done

    done

}



typeset -r c1=Number1

typeset -r c2=Number2

typeset -r c3=Number3

typeset -r c4=All

typeset -r c5=Quit

typeset -ra options=($c1 $c2 $c3 $c4 $c5)

typeset -a answer

typeset -i kk

while true

do

    showmenu

    for kk in "${answer[@]}"

    do

        case $kk in

        1)

            echo 'Number One'

            ;;

        2)

            echo 'Number Two'

            ;;

        3)

            echo 'Number Three'

            ;;

        4)

            echo 'Number One, Two, Three'

            ;;

        5)

            echo 'Program Exit'

            exit 0

            ;;

        esac

    done 

done

Ok will try it out first. Thanks.

This can be done just by wrapping your case block in a for loop and changing one line.

Code:

#!/bin/bash

showmenu () {

    echo "1. Number1"

    echo "2. Number2"

    echo "3. Number3"

    echo "4. All"

    echo "5. Quit"

}



while true ; do

    showmenu

    read choices

    for choice in $choices ; do

        case "$choice" in

            1)

                echo "Number One" ;;

            2)

                echo "Number Two" ;;

            3)

                echo "Number Three" ;;

            4)

                echo "Numbers One, two, three" ;;

            5)

                echo "Exit"

                exit 0 ;;

            *)

                echo "Please enter number ONLY ranging from 1-5!"

                ;;

        esac

    done

done

You can now enter any number of numbers seperated by white space.

Cheers,

EVo2.

Hi all, i am currently trying out this code.

CODE:

showmenu () {
echo "1. Number1"
echo "2. Number2"
echo "3. Number3"
echo "4. All"
echo "5. Quit"
}

while true
do
showmenu
echo "Enter a choice (1-5):"
read choices
clear
echo -e '\E[034;1m'"\033[1mSunOS Audit ...\033[0m"
for choice in $choices
do
case "$choice" in
1)
#clear
echo "Number One"
;;
2)
#clear
echo "Number Two"
;;
3)
#clear
echo "Number Three"
;;
4)
#clear
echo "Number One, Two, Three"
;;
5)
#clear
echo -e '\E[031;1m'"\033[1mPROGRAM CLOSED!\033[0m"
exit 0
;;
*)
echo "INVALID INPUT! PLEASE TRY AGAIN!"
;;
esac
done
echo -e '\E[034;1m'"\033[1mSCAN COMPLETED\033[0m"
exit 0
done

RUN:

1. Number1
2. Number2
3. Number3
4. All
5. Quit
Enter a choice (1-5):
1 32 2

When the code is run, the user will be prompt to enter the choices he or she wants. So right now, i enter the choices "1 32 2". But however choice "32" is actually invalid, so the output will be this:

OUTPUT:
SunOS Audit ...
Number One
INVALID INPUT! PLEASE TRY AGAIN!
Number Two
SCAN COMPLETED

Is it possible that if an invalid choice is being catch, the script will immediately prompt the user to re-enter the choice again? Any idea?

Yes, use "break" to jump out of the for loop, which will send control back to the top of the while loop

Code:

*)

echo "INVALID INPUT! PLEASE TRY AGAIN!"

break

;;

Evo2.