Test for an argument that is a 5-digit integer

Question

I want to check if the argument in my bash

script is a 5 digit integer.

if [ "$1" == [0-9][0-9][0-9][0-9][0-9] ]; then
    echo "first arg is 5 digits"
else
    echo "not 5 digits"

This doesn't work for me, is there an easier way to do this using a function expr

?

+3

linux unix bash regex

user1813619 Jan 25. At 23:39

source to share

3 answers

Anton Kovalenko · Answer 1 · 2013-01-25T23:46:53+0000

See CONDITIONAL EXPRESSIONS

in man bash

:

if [[ "$1" =~ ^[0-9]{5}$ ]]; then
    echo "first arg is 5 digits"
else
    echo "not 5 digits"
fi

michaelmeyer · Answer 2 · 2013-01-26T00:14:34+0000

You must use double parentheses when testing such an expression.

wnoise · Answer 3 · 2013-01-25T23:47:45+0000

case "$1" in
  [0-9][0-9][0-9][0-9][0-9]) echo "first arg is 5 digits" ;;
  *) echo "not 5 digits" ;;
esac