Using Ruby convert numbers to words?

Question

How to convert numbers to words in ruby?

I know there is a gem somewhere. Trying to implement it without a gem. I just need the numbers to words in English for integ

Answer 1

I am not quite sure, if this works for you. Method can be called like this.

n2w(33123) {|i| puts i unless i.to_s.empty?}

Here is the method ( I have not tested it fully. I think it works upto million. Code is ugly, there is a lot of room for re-factoring. )

def n2w(n)
  words_hash = {0=>"zero",1=>"one",2=>"two",3=>"three",4=>"four",5=>"five",6=>"six",7=>"seven",8=>"eight",9=>"nine",
                    10=>"ten",11=>"eleven",12=>"twelve",13=>"thirteen",14=>"fourteen",15=>"fifteen",16=>"sixteen",
                     17=>"seventeen", 18=>"eighteen",19=>"nineteen",
                    20=>"twenty",30=>"thirty",40=>"forty",50=>"fifty",60=>"sixty",70=>"seventy",80=>"eighty",90=>"ninety"}
  scale = {3=>"hundred",4 =>"thousand",6=>"million",9=>"billion"}

  if words_hash.has_key?n
    yield words_hash[n] 
  else
    ns = n.to_s.split(//)
      while ns.size > 0      
        if ns.size == 2
            yield("and")
            yield words_hash[(ns.join.to_i) - (ns.join.to_i)%10]            
            ns.shift
        end
        if ns.size > 4
          yield(words_hash[(ns[0,2].join.to_i) - (ns[0,2].join.to_i) % 10])
        else
          yield(words_hash[ns[0].to_i]) 
        end
        yield(scale[ns.size])
        ns.shift
      end
    end
end