How can I tell whether a variable contains a valid number?

First, you have to define what you mean by "number". The most common case when people ask this seems to be "a non-negative integer, with no leading + sign". Or in other words, a string of all digits. Other times, people want to validate a floating-point input, with optional sign and optional decimal point.

Hand parsing

If you're validating a simple "string of digits", you can do it with a glob:

# Bash / Ksh
if [[ -n $foo && $foo != *[!0123456789]* ]]; then
    printf '"%s" is strictly numeric\n' "$foo"
else
    printf '"%s" has a non-digit somewhere in it or is empty\n' "$foo"
fi >&2

Avoid [0-9] or [[:digit:]] which in some locales and some systems can match characters other than 0123456789.

The same thing can be done in POSIX shells as well, using case:

# POSIX
case $var in
    '')
        printf 'var is empty\n';;
    *[!0123456789]*)
        printf '%s has a non-digit somewhere in it\n' "$var";;
    *)
        printf '%s is strictly numeric\n' "$var";;
esac >&2

Of course, if all you care about is valid vs. invalid, you can combine cases:

# POSIX
case $var in
    '' | *[!0123456789]*)
        printf '%s\n' "$0: $var: invalid digit" >&2; exit 1;;
esac

If you need to allow a leading negative sign, or if want a valid floating-point number or something else more complex, then there are a few possible ways. Standard globs aren't expressive enough to do this, but you can trim off any sign and then compare:

# POSIX
case ${var#[-+]} in   # notice ${var#prefix} substitution to trim sign
    '')
        printf 'var is empty\n';;
    .)
        printf 'var is just a dot\n';;
    *.*.*)
        printf '"%s" has more than one decimal point in it\n' "$var";;
    *[!0123456789.]*)
        printf '"%s" has a non-digit somewhere in it\n' "$var";;
    *)
        printf '"%s" looks like a valid float\n' "$var";;
esac >&2

Or in Bash, we can use extended globs:

# Bash -- extended globs must be enabled explicitly in versions prior to 4.1.
# Check whether the variable is all digits.
shopt -s extglob
[[ $var = +([0123456789]) ]]

A more complex case:

# Bash / ksh
shopt -s extglob

if [[ $foo = @(*[0123456789]*|!([+-]|)) && $foo = ?([+-])*([0123456789])?(.*([0123456789])) ]]; then
  echo 'foo is a floating-point number'
fi

Optionally, case..esac may have been used in shells with extended pattern matching. The leading test of $foo is to ensure that it contains at least one digit, isn't empty, and contains more than just + or - by itself.

If your definition of "a valid number" is even more complex, or if you need a solution that works in legacy Bourne shells, you might prefer to use an external tool's regular expression syntax. Here is a portable version (explained in detail here), using awk (not egrep which is line-based so would be tricked by variables that contain newline characters):

# Bourne
 
if awk -- 'BEGIN {exit !(ARGV[1] ~ /^[-+]?([0123456789]+\.?|[0123456789]*\.[0123456789]+)$/)}' "$foo"; then
    printf '"%s" is a number\n' "$foo"
else
    printf '"%s" is not a number\n' "$foo"
fi

Bash version 3 and above have regular expression support in the [[...]] construct.

# Bash
# The regexp must be stored in a var and expanded for backward compatibility with versions < 3.2

regexp='^[-+]?[0123456789]*(\.[0123456789]*)?$'
if [[ $foo = *[0123456789]* && $foo =~ $regexp ]]; then
    printf '"%s" looks rather like a number\n' "$foo"
else
    printf '"%s" doesn't look particularly numeric to me\n' "$foo"
fi

Using the parsing done by [ and printf (or "using eq")

# fails with ksh
if [ "$foo" -eq "$foo" ] 2>/dev/null; then
 echo "$foo is an integer"
fi

[ parses the variable and interprets it a decimal integer because of the -eq. If the parsing succeeds the test is trivially true; if it fails [ prints an error message that 2>/dev/null hides and sets a status different from 0. However this method fails if the shell is ksh, because ksh evaluates the variable as an arithmetic expression (and that would constitute an arbitrary command injection vulnerability).

Be careful: the following trick with printf (not supported by all shells, and the list of supported float representations varies with the shell as well; not to mention the command injection vulnerability in ksh or zsh)

if printf %f "$foo" >/dev/null 2>&1; then
  printf '"%s" is a float\n' "$foo"
fi

is broken: about the arguments of the a, A, e, E, f, F, g, or G format modifiers, POSIX specifies that if the leading character is a single-quote or double-quote, the value shall be the numeric value in the underlying codeset of the character following the single-quote or double-quote. Hence this fails when foo expands to a string with a leading single-quote or double-quote: the previous command will happily validate the string as a float. It also returns 0 when foo expands to a number with a leading 0x, which is a valid number in a shell script but may not work elsewhere.

You can use %d to parse an integer. Take care that the parsing might be (is supposed to be?) locale-dependent.

BashFAQ/054 (last edited 2020-10-15 10:20:40 by StephaneChazelas)