Bash if variable is an integer

Question

I'm trying to write an if statement in bash that will exit if the variable supplied is not an integer. I will eventually be nesting this if statement within a while loop.

When I do run this I am getting an syntax error.

#!/bin/bash
if [ $f1 != ^[0-9]+$ ]
    then 
        exit 1 
    fi

by integer, are you including negative integers or only positive? — LinkBerest - SO sold our work
– LinkBerest - SO sold our work, Commented Apr 24, 2015 at 15:50

David C. Rankin · Accepted Answer · 2015-04-24 16:24:13Z

5

I have always like the integer test using the equality test construct:

[ $var -eq $var 2>/dev/null ] || exit 1

If var is not an integer, the equality fails due to the error generated. It is also POSIX compliant as it doesn't rely on character classes or the bash [[ construct.

answered Apr 24, 2015 at 16:24

David C. Rankin

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Comments

fedorqui · Accepted Answer · 2015-04-24 16:19:09Z

2

You better negate the condition like this:

if [[ ! "$f1" =~ ^[0-9]+$ ]]; then 
  exit 1
fi

note the [[ and ]] syntax for the regular expressions, together with ! to negate it. Then, we use =~ for regexs.

Test

$ r=23a
$ [[ ! "$r" =~ ^[0-9]+$ ]] && echo "no digit" || echo "digit"
no digit
$ r=23
$ [[ ! "$r" =~ ^[0-9]+$ ]] && echo "no digit" || echo "digit"
digit

edited Apr 24, 2015 at 16:19

answered Apr 24, 2015 at 15:48

fedorqui

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3 Comments

thisguy Over a year ago

thank you very much can you explain to me why this is important and what this command do to make this work correctly ? note the [[ and ]] syntax, together with ! "$var" =~.

LinkBerest - SO sold our work Over a year ago

In case you want negative integers just add a -? (optional '-') to the start of the regex as such: [[ ! "$f1" =~ ^-?[0-9]+$ ]]

fedorqui Over a year ago

@thisguy this is well explained in The conditional expression. Specifically when it talks about <STRING> != <PATTERN> as <STRING> is checked against the pattern <PATTERN> - TRUE on no match.

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