I want to write program to check whether the given input is number or not. Can anyone help me?
if [ $Number -ne 0 -o $Number -eq 0 2>/dev/null ]
then ...
and what -o stands for in above command?
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2 Answers
-o is the or operator. So your test check if your number is not equal to 0 or if it's equals to 0.
(so it should always return true).
To check if it's a number, you could use a regexp: this should be working:
[[ "$number" =~ ^[0-9]+$ ]]
To see the list of all available flags, you should look at man test
Details:
[[is a extended bash test command. It supports more operator thantest/[.=~compare the first argument again a regular expression^[0-9]+$is the regular expression.^is an anchor for the start of the string,$for the end.[0-9]means any char between 0 and 9, and+is for the repetition of the latest pattern
2 Comments
Srikanth Power
thank you what u gave is working fine for my programm.I am new to linux can you please explain the above command given by you?
AdrieanKhisbe
update the question with details :)
The POSIX way to do this (for integers; floats are more complicated) is
case $number in
(*[^0-9]*) printf '%s\n' "not a number";;
() printf '%s\n' "empty";;
(*) printf '%s\n' "a number";;
esac
This works for any Bourne shell, unlike the [[ ... ]] construct, which is non-portable.
