exact year to select with bash script

Question

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?

In my opinion, this can be removed: `-z $y || $y =~ [A-Za-z]+ ||` — Cyrus, Mar 26 '17 at 17:16

score 3 · Answer 1 · edited May 23 '17 at 12:17

3

Replace

[0-9]{4}

by

^[0-9]{4}$

to match only exact four numbers.

See: The Stack Overflow Regular Expressions FAQ

edited May 23 '17 at 12:17

Community

1
1

answered Mar 26 '17 at 17:11

Cyrus

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2

Or use a glob pattern with the == operator: `$y == [0-9][0-9][0-9][0-9]` – glenn jackman Mar 26 '17 at 17:15
Yes, that is possible, too. – Cyrus Mar 26 '17 at 17:16

score 0 · Answer 2 · answered Mar 26 '17 at 17:21

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.

exact year to select with bash script

2 Answers2