CODEWITHSUNDEEP
bash
3kstc
3kstc

Reputation: 1966

How to check if a string only contains digits/numerical characters

How can I check if MyVar contains only digits with an if statement with BASH. By digits I am referring to 0-9.

ie:

if [[ $MyVar does contain digits ]]  <-- How can I check if MyVar is just contains numbers
then
 do some maths with $MyVar
else
 do a different thing
fi

Upvotes: 28

Views: 50016

Answers (6)

Bruno
Bruno

Reputation: 621

For Bash, another option, maybe less expensive than regex match (I did not make performance tests), is to use pattern substitution to replace all digits with empty string, and test if the resulting string is empty.

if [[ -z ${var//[0-9]} ]]; then      # or ${var//[[:digit:]]}
   echo "only digits"
else
   echo "not only digits"
fi

Upvotes: 3

MrPotatoHead
MrPotatoHead

Reputation: 1144

Grep expression solution without [[ ]], for wider shell compatability:

if [ "$(echo $var | grep -E "^[0-9]{1,}$")" ]; then echo "digits only"; then
    echo "var contains digits only"
else
    echo "var contains digits and/or other characters"
fi

Upvotes: 1

Abhishek J
Abhishek J

Reputation: 111

Simple! just use expression option in grep The below solution echo's the variable and checks if it contains only digits

if [[ $(echo $var | grep -E "^[[:digit:]]{1,}$") ]]
then
    echo "var contains only digits"
else
    echo "var contains other characters apart from digits"
fi

Upvotes: 1

David C. Rankin
David C. Rankin

Reputation: 84561

If you would like to test in a POSIX compliant way, you can use either:

expr string : regex        ## returns length of string if both sides match

or

expr match string regex    ## behaves the same

For example to test if $myvar is all digits:

[ $(expr "x$myvar" : "x[0-9]*$") -gt 0 ] && echo "all digits"

Note: 'x' prepended to the variable and expression to protect against test of empty-string throwing error. To use the length returned by the test, don't forget to subtract 1 which represents the 'x'.

In if-then-else form, here is a short script that tests whether the first argument to the script contains all digits:

#!/bin/sh

len=$(expr "x$1" : "x[0-9]*$")  ## test returns length if $1 all digits
let len=len-1                   ## subtract 1 to compensate for 'x'

if [ $len -gt 0 ]; then         ## test if $len -gt 0 - if so, all digits
    printf "\n '%s' : all digits, length: %d chars\n" "$1" $len
else
    printf "\n '%s' : containes characters other than [0-9]\n" "$1"
fi

Example Output

$ sh testdigits.sh 265891

 '265891' : all digits, length: 6 chars

$ sh testdigits.sh 265891t

 '265891t' : contains characters other than [0-9]

The bash regular expression test [[ $var =~ ^[0-9]+$ ]] is fine, I use it, but it is a bashism (limited to the bash shell). If you are concerned with portability, the POSIX test will work in any POSIX compliant shell.

Upvotes: 11

John1024
John1024

Reputation: 113844

[[ $myvar =~ [^[:digit:]] ]] || echo All Digits

Or, if you like the if-then form:

if [[ $myvar =~ [^[:digit:]] ]]
then
    echo Has some nondigits
else
    echo all digits
fi

In olden times, we would have used [0-9]. Such forms are not unicode safe. The modern unicode-safe replacement is [:digit:].

Upvotes: 18

giliev
giliev

Reputation: 3058

Here it is:

#!/bin/bash
if [[ $1 =~ ^[0-9]+$ ]]
then
    echo "ok"
else
    echo "no"
fi

It prints ok if the first argument contains only digits and no otherwise. You could call it with: ./yourFileName.sh inputValue

Upvotes: 41

Related Questions