Bash script

spartiati · 01-28-2010, 10:23 AM

I am fairly new to bash scripting, and I am trying to test a variable against a number range (1-3).

This is what I have used so far:
The user enters a value, then

while [ "$PROV" != "1" ] && [ "$PROV" != "2" ] && [ "$PROV" != "3" ] && [ $COUNTER -lt 4 ] || [ -z "$PROV" ] && [ $COUNTER -lt 4 ]; do
clear
echo
echo "You did not enter an appropriate choice. Please enter 1, 2, or 3."
echo "1)PROV1"
echo "2)Prov2"
echo "3)Prov3"
echo
echo "Enter Choice:"
read -e PROV
echo
let COUNTER++
done

Is there a cleaner way to do this?
Something like:
while [ "$PROV" != "1-3" ]&& [ $COUNTER -lt 4 ] || [ -z "$PROV" ] && [ $COUNTER -lt 4 ]; do

Thank you.

EricTRA · 01-28-2010, 10:37 AM

Hello and Welcome to LinuxQuestions,

Why not use a case statement? That's a lot easier to read and understand.

Code:

read PROV
case $PROV in
[1-3])
   echo "You entered a correct value"
   exit
;;
*)
   echo "You entered an incorrect value"
   exit
;;
esac

Here's a great Bash guide: Bash Guide for Beginners

Kind regards,

Eric

chrism01 · 01-29-2010, 12:26 AM

Also, you're using the wrong comparison operator http://tldp.org/LDP/abs/html/comparison-ops.html

konsolebox · 01-29-2010, 01:35 AM

Following EricTRA's suggestion, here's my version:

Code:

#!/bin/bash

for (( COUNTER=0;; )); do
    read PROV

    if [[ $PROV == [1-3] ]]; then
        echo "You entered a correct value."
        break
    fi

    (( ++COUNTER ))

    if [[ COUNTER -eq 3 ]]; then
        echo "You entered too many incorrect values."
        break
    else
        echo "You entered an incorrect value."
    fi
done