How to test if a string is basically an integer in quotes using Ruby

Question

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

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  

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.

"12".match(/^(\d)+$/)      # true
"1.2".match(/^(\d)+$/)     # false
"dfs2".match(/^(\d)+$/)    # false
"13422".match(/^(\d)+$/)   # true