Check whether input is number or not in bash

Question

I want to write program to check whether the given input is number or not. Can anyone help me?

 if [ $Number -ne 0 -o $Number -eq 0 2>/dev/null ]
 then ...

and what -o stands for in above command?

AdrieanKhisbe · Accepted Answer · 2015-03-10T09:06:30.993

8

-o is the or operator. So your test check if your number is not equal to 0 or if it's equals to 0.

(so it should always return true).

To check if it's a number, you could use a regexp: this should be working:

[[ "$number" =~ ^[0-9]+$ ]]

To see the list of all available flags, you should look at man test

Details:

[[ is a extended bash test command. It supports more operator than test/[.
=~ compare the first argument again a regular expression
^[0-9]+$ is the regular expression. ^ is an anchor for the start of the string, $ for the end. [0-9] means any char between 0 and 9, and + is for the repetition of the latest pattern

edited Mar 10 '15 at 09:06

answered Mar 10 '15 at 06:45

AdrieanKhisbe

thank you what u gave is working fine for my programm.I am new to linux can you please explain the above command given by you? – Srikanth Power Mar 10 '15 at 09:01
update the question with details :) – AdrieanKhisbe Mar 10 '15 at 09:06

score 3 · Answer 2 · answered Mar 10 '15 at 07:57

The POSIX way to do this (for integers; floats are more complicated) is

 case $number in
   (*[^0-9]*) printf '%s\n' "not a number";;
   ()         printf '%s\n' "empty";;
   (*)        printf '%s\n' "a number";;
 esac

This works for any Bourne shell, unlike the [[ ... ]] construct, which is non-portable.

2 Answers2