How to test if a string is basically an integer in quotes using Ruby
Solution 1:
Well, here's the easy way:
class String
def is_integer?
self.to_i.to_s == self
end
end
>> "12".is_integer?
=> true
>> "blah".is_integer?
=> false
I don't agree with the solutions that provoke an exception to convert the string - exceptions are not control flow, and you might as well do it the right way. That said, my solution above doesn't deal with non-base-10 integers. So here's the way to do with without resorting to exceptions:
class String
def integer?
[ # In descending order of likeliness:
/^[-+]?[1-9]([0-9]*)?$/, # decimal
/^0[0-7]+$/, # octal
/^0x[0-9A-Fa-f]+$/, # hexadecimal
/^0b[01]+$/ # binary
].each do |match_pattern|
return true if self =~ match_pattern
end
return false
end
end
Solution 2:
You can use regular expressions. Here is the function with @janm's suggestions.
class String
def is_i?
!!(self =~ /\A[-+]?[0-9]+\z/)
end
end
An edited version according to comment from @wich:
class String
def is_i?
/\A[-+]?\d+\z/ === self
end
end
In case you only need to check positive numbers
if !/\A\d+\z/.match(string_to_check)
#Is not a positive number
else
#Is all good ..continue
end
Solution 3:
You can use Integer(str)
and see if it raises:
def is_num?(str)
!!Integer(str)
rescue ArgumentError, TypeError
false
end
It should be pointed out that while this does return true for "01"
, it does not for "09"
, simply because 09
would not be a valid integer literal. If that's not the behaviour you want, you can add 10
as a second argument to Integer
, so the number is always interpreted as base 10.