CODEWITHSUNDEEP
bashif-statement
swa056
swa056

Reputation: 95

Create variable within a range of numbers except a list of numbers

I have the following variable with a list of numbers

vlist="1 13 20 21 22 24 25 28 240 131 133 136 138 224"

In the next loop I want to input a number between 1 - 250 except the numbers in vlist

while :; do
echo -en 'Number : ' ; read -n3 cvip ; echo
    [[ $cvip =~ ^[0-9]+$ ]] || { echo -e '\nSorry input a number between 1 and 239 except '$vlist'\n' ; continue; }
    cvip=$(expr ${cvip} + 0)
    if ((cvip >= 1 && cvip <= 250)); then break ; else echo -e '\nNumber out of Range, input a number between 1 and 239 except '$vlist'\n' ; fi
done

Ηow can I enter the list exception inside the if statement range

Upvotes: 0

Views: 62

Answers (1)

Shawn
Shawn

Reputation: 52449

If using bash, one approach is to store the bad numbers as keys in an associative array and see if that particular key exists when validating a number:

#!/usr/bin/env bash

vlist="1 13 20 21 22 24 25 28 240 131 133 136 138 224"
declare -A nums
for num in $vlist; do
    nums[$num]=1
done

while true; do
    while read -p "Number: " -r num; do
        if [[ ! $num =~ ^[[:digit:]]+$ ]]; then
            echo "Input an integer."
        elif [[ ${nums[$num]:-0} -eq 1 ]]; then
            echo "Input a number between 1 and 250 except '$vlist'"
        elif [[ $num -lt 1 || $num -gt 250 ]]; then
            echo "Input an integer between 1 and 250"
        else
            break
        fi
    done
    printf "You successfully inputted %d\n" "$num"
done

The important bit is ${nums[$num]:-0}, which expands to the value of that element of the associative array if that key exists and is not null, or 0 otherwise. As noted by Glenn in a comment, in bash 4.3 or newer, [[ -v nums[$num] ]] works too for testing to see if a given key exists, so [[ ${nums[$num]:-0} -eq 1 ]] in the above could be replaced with it.

Upvotes: 1

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