Test if a variable has more that 4 digits in bash script

Question

i want to test if a variable has more than 4 digits something like this

#!/bin/bash
if [ $input has more than 4 digits ]; then 
     echo "  * Please only 4 digits" >&2
     echo""
else
   the other option
fi

Answer 1

If you care about the number of digits (and not the numerical value), you could match against a regex in Bash/Ksh/Zsh ^{(* see footnote on [[:digit:]])}:

#!/bin/bash
input=$1
re='^[[:digit:]]{1,4}$'
if [[ $input =~ $re ]]; then
    echo "'$input' contains 1 to 4 digits (and nothing else)"
else
    echo "'$input' contains something else"
fi

Or e.g. [[ $input =~ ^[[:digit:]]{5,}$ ]] to check for "5 or more digits (and nothing else)", etc.

---

Or in a pure POSIX shell, where you have to use case for the pattern match:

#!/bin/sh
input=$1
case $input in 
    *[![:digit:]]*) onlydigits=0;; # contains non-digits
    *[[:digit:]]*)  onlydigits=1;; # at least one digit
    *)              onlydigits=0;; # empty
esac
if [ $onlydigits = 0 ]; then
    echo "'$input' is empty or contains something other than digits"
elif [ "${#input}" -le 4 ]; then
    echo "'$input' contains 1 to 4 digits (and nothing else)"
else
    echo "'$input' contains 5 or more digits (but nothing else)"
fi

(You could put all the logic inside the case, but nesting an if there is somewhat ugly, IMO.)

---

_{Note that [[:digit:]] should match whatever the current locale's idea of "digits" is. That might or might not be more than the ASCII digits 0123456789. On my system, [[:digit:]] does not match e.g. ⁴ (superscript four, U+2074), but [0-9] does. Matching other "digits" might be a problem, esp. if you do arithmetic on the number in the shell. So, if you want to be stricter, use [0123456789] to accept just the ASCII digits.}

Answer 2

Here assuming you mean ASCII decimal digits only and not other sorts of decimal or non-decimal digits.

shopt -s extglob # enables a subset of ksh extended globs including *(...),
                 # +(...) and ?(...) but unfortunately not {4}(...)
d='[0123456789]' nd='[^0123456789]'
case $input in
  ( $d$d$d$d+($d)     ) echo made of more than 4 digits;;
  ( $d$d$d$d$d  ) echo contains more than 4 digits;;
  ( ""                ) echo empty;;
  ( ($nd)            ) echo does not contain any digit;;
  ( $nd*             ) echo no more than 4 digits but also contains non-digits;;
  ( $d?($d)?($d)?($d) ) echo made of 1 to 4 digits;;
  ( *                 ) echo should not be reached;;
esac

Beware that in bash and depending on the system and locale, [0-9] and [[:digit:]] may match a lot more than just 0123456789 so those should not be used for input validation (more on that in that answer to a different question here for instance).

Also beware that bash pattern matching works in very-surprising ways in multi-byte locales.

You'll find that for instance in a zh_CN.gb18030 Chinese locale, on input='1-©©' it will return no more than 4 digits but also contains non-digits as expected, but if you append a single 0x80 byte (input='1-©©'$'\x80'), it will return contains more than 4 digits.

It's for this kind of reason (and the fact that pattern matching has been known to have bugs in corner cases in many shells) that for input validation, it's better to use a positive matching where possible for the things you accept (rather than negative match for the things to reject)¹ hence the $d?($d)?($d)?($d) above even though it shouldn't be necessary as in theory at least, anything else should have been matched by earlier patterns.

---

^{¹ as an exception to that, one may need to consider the Bourne and Korn shell's misfeature whereby case $input in [x]) echo yes; esac matches on x but also on [x]!}

Answer 3

I'd do

#!/usr/bin/env bash
die () { echo "$*" >&2; exit 1; }
input=$1
[[ $input == +([[:digit:]]) ]] || die "only digits please"
(( input <= 9999 ))            || die "no more than 4 digits please"
echo "ok: $input"

Answer 4

If you want to examine the number of characters a variable, you can do this...

var="foo bar"
echo "var contains ${#var} characters"
Result:
var contains 7 characters

Answer 5

This is another way:

#!/bin/bash
if test -z "$1"
then
    echo no digit supplied
elif grep -qE '[[:digit:]]{5}' <<< "$1"
then
    echo too many digits supplied
else
    echo number of digits ok
fi