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Revision 2 as of 2007-05-07 18:02:07
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Editor: GreyCat
Comment: [0-9]* before the decimal, not [0-9]+, so we can have .5 as a number. This requires checking non-blank-ness.
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Editor: cpe-74-65-28-251
Comment: Fixed broken escaping
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if [[ $foo && $foo =~ ^[-+]?[0-9]*\(\.[0-9]+\)?$ ]]; then if [[ $foo && $foo =~ ^[-+]?[0-9]*(\.[0-9]+)?$ ]]; then


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 seems to be that, when people ask this, they actually mean "a non-negative integer, with no leading + sign".

if [[ $foo = *[^0-9]* ]]; then
   echo "'$foo' has a non-digit somewhere in it"
   echo "'$foo' is strictly numeric"

This can be done in Korn and legacy Bourne shells as well, using case:

case "$foo" in
    *[!0-9]*) echo "'$foo' has a non-digit somewhere in it" ;;
    *) echo "'$foo' is strictly numeric" ;;

If what you actually mean is "a valid floating-point number" or something else more complex, then you might prefer to use a regular expression. Bash version 3 and above have regular expression support in the [[ command:

if [[ $foo && $foo =~ ^[-+]?[0-9]*(\.[0-9]+)?$ ]]; then
    echo "'$foo' looks rather like a number"
    echo "'$foo' doesn't look particularly numeric to me"

If you don't have bash version 3, then you would use egrep:

if test "$foo" && echo "$foo" | egrep '^[-+]?[0-9]*(\.[0-9]+)?$' >/dev/null; then
    echo "'$foo' might be a number"
    echo "'$foo' might not be a number"

Note that the parentheses in the egrep regular expression don't require backslashes in front of them, whereas the ones in the bash3 command do. Also, the leading test of "$foo" (in both versions) is to ensure that it is not an empty string.

BashFAQ/054 (last edited 2022-04-19 05:23:33 by emanuele6)