Bash script. Accept integer only if is from range

Question

I have bash script with basic arithmetic operations - Addition, Subtraction, Division and Multiplication.

    #! bin/bash

input="yes"
while [[ $input = "yes" ]]
do

    PS3="Press 1 for Addition, 2 for subtraction, 3 for multiplication and 4 for division: "
    select math in Addition Subtraction Multiplication Division
    do
        case "$math" in
        Addition)
            echo "Enter first no:"
            read num1
            echo "Enter second no:"
            read num2
            result=`expr $num1 + $num2`
            echo Answer: $result
            break
        ;;
        Subtraction)
            echo "Enter first no:"
            read num1
            echo "Enter second no:"
            read num2
            result=`expr $num1 - $num2`
            echo Answer: $result
            break
        ;;
        Multiplication)
            echo "Enter first no:"
            read num1
            echo "Enter second no:"
            read num2
            result=`expr $num1 * $num2`
            echo Answer: $result
            break
        ;;
        Division)
            echo "Enter first no:"
            read num1
            echo "Enter second no:"
            read num2
            result=$(expr "scale=2; $num1/$num2" | bc)
            echo Answer = $result
            break
        ;;
        *)
            echo Choose 1 to 4 only!!!!
            break
        ;;
    esac
    done

done

How to make that values for @num1 and @num2 are accepted only if they are numbers in certain range. For example 0 to 10. So if I enter for $num1 or $num2 lets say 500 there will be message to enter valid value?

What have you tried? You should be able to come up with a test that you can use to break out of the case statement with a message. This might be a good candidate for a function. — user1531971
– user1531971, Commented May 7, 2018 at 18:27
while (( $num1 < 0 || $num1 > 10 )); do read -p'Number must be between 0 and 10.' num1; ... done — 0x5453
– 0x5453, Commented May 7, 2018 at 18:50
expr is antiquated pre-POSIX syntax. Use result=$((num1 * num2)) for math in modern POSIX-compliant shells. The expr syntax requires forking a subshell, setting up a pipeline to read that subshell's output, executing an external binary, read()ing its output, wait()ing to collect its exit status, etc; the native syntax just does the math internally to the bash executable. — Charles Duffy
– Charles Duffy, Commented May 7, 2018 at 19:24

PesaThe · Accepted Answer · 2018-05-07 19:30:48Z

2

You can create a simple function to get a number in range:

get_number() {
    local lo=$1 up=$2 text=${3:-Enter a number: } num      
    shopt -s extglob

    until   
        read -p "$text" num
        [[ $num = ?(-)+([0-9]) ]] && (( $lo <= 10#$num && 10#$num <= $up ))
    do
        echo "Invalid input!" >&2
    done
    echo "$((10#$num))"
}

num1=$(get_number 10 15 "Enter first number: ")
num2=$(get_number -10 20) # use default prompt

Works for integers only. You might also consider inputting the numbers before the case command to avoid redundant code.

edited May 7, 2018 at 19:30

answered May 7, 2018 at 18:39

PesaThe

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4 Comments

glenn jackman Over a year ago

and shopt -s extglob is required.

PesaThe Over a year ago

@glennjackman thanks for the tip. Is extglob required? Perhaps in older versions but quoting manual: ...matched according to the rules described below in Pattern Matching, as if the extglob shell option were enabled..

Charles Duffy Over a year ago

That behavior is indeed version-dependent; safer to enable the flag for compatibility.

PesaThe Over a year ago

@CharlesDuffy I see. Thanks.

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