CODEWITHSUNDEEP
linuxbash
Jim Jonas
Jim Jonas

Reputation: 5

Check whether a number is single, double, or triple digit using Bash

I am trying to implement a Bash script that checks whether the number given by the user is single/2-digit/3-digit. I can only use case..esac statement. The following program works fine for single digit but the reset of the cases doesn't execute.

echo -e "\n-----NUMBER CHECK-----"
echo -e -n "\nEnter a number: "
read num

case $num in 
  [0-9])
    echo -e "\nSINGLE DIGIT NUMBER!"
  ;;

  [10-99])
    echo -e "\nDOUBLE DIGIT NUMBER!"
  ;;

  [100-999])
    echo -e "\nTRIPLE DIGIT NUMBER!"
  ;;
  
  *)
    echo -e "\nInvalid Input!"
  ;;

esac

Upvotes: 0

Views: 4342

Answers (4)

fpmurphy
fpmurphy

Reputation: 2537

An alternative but hacked solution:

echo -e "\n-----NUMBER CHECK-----"
echo -e -n "\nEnter a number: "
read num

case $num in
    *[!0-9]*|????*) good=0 ;;
    *) good=1;
esac

case 1 in
  $((good == 1 && num <= 9)))
    echo -e "\nSINGLE DIGIT NUMBER!"
  ;;

  $((good == 1 && num <= 99)))
    echo -e "\nDOUBLE DIGIT NUMBER!"
  ;;

  $((good == 1 && num <= 999)))
    echo -e "\nTRIPLE DIGIT NUMBER!"
  ;;

  *)
    echo -e "\nInvalid Input!"
  ;;

esac

Upvotes: 0

Jetchisel
Jetchisel

Reputation: 7801

Something like this.

#!/usr/bin/env bash

printf '\n-----NUMBER CHECK-----\n\n'

read -rp 'Enter a number: ' num

case $num in
  [0-9])
    printf '\nSINGLE DIGIT NUMBER!'
    ;;
  [0-9][0-9])
    printf '\nDOUBLE DIGIT NUMBER!'
    ;;
  [0-9][0-9][0-9])
    printf '\nTRIPLE DIGIT NUMBER!'
    ;;
  *)
    printf >&2 '\nINVALID INPUT!'
    ;;
esac

An alternative is to use a regex pattern matching using the test =~ operator if your version of bash supports it.

#!/usr/bin/env bash

printf '\n-----NUMBER CHECK-----\n\n'

read -rp 'Enter a number: ' num

if [[ $num =~ ^[[:digit:]]{1}$ ]]; then
  printf '\nSINGLE DIGIT NUMBER!'
elif [[ $num =~ ^[[:digit:]]{2}$ ]]; then
  printf '\nDOUBLE DIGIT NUMBER!'
elif [[ $num =~ ^[[:digit:]]{3}$ ]]; then
  printf '\nTRIPLE DIGIT NUMBER!'
else
  printf >&2 '\nINVALID INPUT!'
fi

Upvotes: 1

oguz ismail
oguz ismail

Reputation: 50775

I'd validate the input first.

case $num in
*[!0-9]*|????*) ;; # invalid input
?) ;;              # single digit
??) ;;             # double digit
???) ;;            # triple digit
esac

Upvotes: 4

Socowi
Socowi

Reputation: 27225

[] is a character class. The - for specifying ranges only handles characters, not integers. [100-999] is a character class with …

  • the characters 1, 0
  • the range 0-9
  • and the characters 9 and 9

… so basically [0-9].

As pointed out by Jetchisel, use [0-9][0-9] and [0-9][0-9][0-9].

Upvotes: 2

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