CODEWITHSUNDEEP
linuxbashunixsh
Kshitij8097
Kshitij8097

Reputation: 55

exact year to select with bash script

I am working on following script which throws an error if the year is not in digits with exact 4 digits.

so far, I have below code but the only problem is it accepting more digits than 4 which I want to limit it to 4 digits only.

#!/bin/bash

read -p "Enter year " y
while [[ -z $y || $y =~ [A-Za-z]+ || ! $y =~ [0-9]{4} ]]; do
            read -p "Enter the year in a valid format [yyyy] " y
done
echo "Selected year is: $y"

output -

 Enter the year in a valid format [yyyy] vrgege
 Enter the year in a valid format [yyyy] vegter
 Enter the year in a valid format [yyyy] 123
 Enter the year in a valid format [yyyy] 12345
 Selected year is: 12345

It not accepting less than 4 digits but it accepting more than 4 digits.

Can anyone suggest, how to put a condition to get exact only 4 digits?

Upvotes: 0

Views: 75

Answers (2)

Jens
Jens

Reputation: 72732

Without bashisms, for portability, I suggest

case $y in
([0-9][0-9][0-9][0-9])
   printf "A four digit year, OK\n"
   ;;
([0-9][0-9][0-9])
   printf "A three digit year, OK\n"
   ;;
([0-9][0-9])
   printf "A two digit year, OK\n"
   ;;
([0-9])
   printf "A one digit year, OK\n"
   ;;
(*)
   printf "NOT OK\n"
esac

You can also combine these with (...|...|...|...) if you wish.

Upvotes: 0

Cyrus
Cyrus

Reputation: 88889

Replace

[0-9]{4}

by

^[0-9]{4}$

to match only exact four numbers.

See: The Stack Overflow Regular Expressions FAQ

Upvotes: 3

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