[0-9]* before the decimal, not [0-9]+, so we can have .5 as a number. This requires checking non-blank-ness.
Fixed broken escaping
|Deletions are marked like this.||Additions are marked like this.|
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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" else echo "'$foo' is strictly numeric" fi
This can be done in Korn and legacy Bourne shells as well, using case:
case "$foo" in *[!0-9]*) echo "'$foo' has a non-digit somewhere in it" ;; *) echo "'$foo' is strictly numeric" ;; esac
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 "'$foo' looks rather like a number" else echo "'$foo' doesn't look particularly numeric to me" fi
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" else echo "'$foo' might not be a number" fi
Note that the parentheses in the egrep regular expression don't require backslashes in front of them, whereas the ones in the bash3 command do. Also, the leading test of "$foo" (in both versions) is to ensure that it is not an empty string.