one RE does it all
eq method added
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|=== Using the 'eq' operator ===
is a number
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.
If you're validating a simple "string of digits", you can do it with a glob:
# Bash if [[ $foo != *[!0-9]* ]]; then echo "'$foo' is strictly numeric" else echo "'$foo' has a non-digit somewhere in it" fi
The same thing can be done in Korn and POSIX shells as well, using case:
# ksh, POSIX case "$foo" in *[!0-9]*) echo "'$foo' has a non-digit somewhere in it" ;; *) echo "'$foo' is strictly numeric" ;; 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 we can use extended globs:
# Bash -- extended globs must be enabled. # Check whether the variable is all digits. shopt -s extglob [[ $var == +([0-9]) ]]
A more complex case:
# Bash shopt -s extglob [[ $foo = *[0-9]* && $foo = ?([+-])*([0-9])?(.*([0-9])) ]] && echo "foo is a floating-point number"
The leading test of $foo is to ensure that it contains at least one digit. The extended glob, by itself, would match the empty string, or a lone + or -, which may not be desirable behavior.
Korn shell has extended globs enabled by default, but lacks [[, so we must use case to do the glob-matching:
# Korn case $foo in *[0-9]*) case $foo in ?([+-])*([0-9])?(.*([0-9]))) echo "foo is a number";; esac;; esac
Note that this uses the same extended glob as the Bash example before it; the third closing parenthesis at the end of it is actually part of the case syntax.
If your definition of "a valid number" is even more complex, or if you need a solution that works in legacy Bourne shells, you might prefer to use an external tool's regular expression syntax. Here is a portable version (explained in detail here), using egrep:
# Bourne if echo "$foo" | egrep '^[-+]?([0-9]+\.?|[0-9]*\.[0-9]+)$' >/dev/null then echo "'$foo' is a number" else echo "'$foo' is not a number" fi
Bash version 3 and above have regular expression support in the [[ command. Due to bugs and changes in the implementation of the =~ feature throughout bash 3.x, we do not recommend using it, but people do it anyway, so we have to maintain this example (and keep restoring this warning, too, when people delete it):
# Bash # Put the RE in a var for backward compatibility with versions <3.2 regexp='^[-+]?[0-9]*(\.[0-9]*)?$' if [[ $foo = *[0-9]* && $foo =~ $var ]]; then echo "'$foo' looks rather like a number" else echo "'$foo' doesn't look particularly numeric to me" fi
Using the parsing done by [ and printf
# fails with ksh if [ "$foo" -eq "$foo" ] 2>/dev/null;then echo "$foo is an integer" fi
[ parses the variable and interprets it as in integer because of the -eq. If the parsing succeds 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.
You can use a similar trick with printf:
# POSIX if printf "%f" "$foo" >/dev/null 2>&1; then echo "$foo is a float" fi
You can use %d to parse an integer. Take care that the parsing might be (is supposed to be?) locale-dependent.
Using the integer type
If you just want to guarantee ahead of time that a variable contains an integer, without actually checking, you can give the variable the "integer" attribute.
# Bash declare -i foo foo=-10+1; echo "$foo" # prints -9 foo="hello"; echo "$foo" # the value of the variable "hello" is evaluated; if unset, foo is 0 foo="Some random string" # results in an error.
Any value assigned to a variable with the integer attribute set is evaluated as an arithmetic expression just like inside $(( )). Bash will raise an error if you try to assign an invalid arithmetic expression.
In Bash and ksh93, if a variable which has been declared integer is used in a read command, the user's input is treated as an arithmetic expression, as with assignment. In particular, if the user types an identifier, the variable will be set to the value of the variable with that name, and read will give no other indication of a problem.
# Bash (and ksh93, if you replace declare with typeset) $ declare -i foo $ read foo hello $ echo $foo # prints 0; 'hello' is unset, so is treated as 0 for arithmetic purposes $ hello=5 $ read foo # user types hello again hello $ echo $foo # prints 5, the value of 'hello' as an arithmetic expression
Pretty useless if you want to read only integers.
In the older Korn shell (ksh88), if a variable is declared integer and used in a read command, and the user types an invalid integer, the shell complains, the read command returns an error status, and the value of the variable is unchanged.
# ksh88 $ typeset -i foo $ foo=42 $ read foo hello ksh: hello: bad number $ echo $? 1 $ echo $foo 42
Using the 'eq' operator
# Bash dev null ; is a number isn