如何在不使用任何宝石的情况下将数字转换为红宝石中的单词？

Question

我有一个数字说100我希望它在word.similary中转换成一百个浮点数说0.0是零和3.4到三点四

Answer 1

有许多宝石可以将数字转换为单词：

• 的 gem 'to_words'
• 的 humanize
• 的 numbers_and_words

但是，由于你不想使用宝石，我发现了类似的东西（没有任何宝石）：

class Fixnum
 def english_word
  @h = { 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=>"Fourty",50=>"Fifty",60=>"Sixty",70=>"Seventy", 80=>"Eighty",90=>"Ninty" }
  @i=0
  @array=[]
  @result=""a
  if self > 99
    str_num=self.to_s #@num.to_s
    str_num_len=str_num.length
    str_full_num=str_num.insert(0,"0"*(11-str_num_len))
    str_full_num=str_num.insert(8,"0")
    str_full_num.scan(/../) { |x|  @array<<x }
    6.times do
    self.def_calc
    @i+=1
    end
  else
     if self > 9
        puts (self.proc_double_dig((self/10)*10)) + (self.proc_single_dig(self%10))
     else
       if self > 0
         puts self.proc_single_dig(self)
       else
         return "AMOUNT NOT KNOWN or NILL"
       end
     end
  end
  end

  def def_calc
    case @i
      when 0
        str=self.proc_unit(@array[@i])
        if (str.scan(/\w+/)).length!=0
             then str=str+ "hundred & "
              @result=@result+str
        end
      when 1
        str=self.proc_unit(@array[@i])
        if (str.scan(/\w+/)).length!=0
             then str=str+ " Crore, "
             @result=@result+str
        end
      when 2
        str=self.proc_unit(@array[@i])
        if (str.scan(/\w+/)).length!=0
             then str=str+ " Lakh, "
             @result=@result+str
        end
      when 3
        str=self.proc_unit(@array[@i])
        if (str.scan(/\w+/)).length!=0
             then str=str+ " Thousand, "
             @result=@result+str
        end
      when 4
        str=self.proc_unit(@array[@i])
        if (str.scan(/\w+/)).length!=0
             then str=str+ " Hundred, "
             @result=@result+str
        end
      when 5
        str=self.proc_unit(@array[@i])
        if (str.scan(/\w+/)).length!=0
             then str=str+ ". "
             @result=@result+str
        end
        print @result.sub(/..$/,"")
    else
   end
  end

  def proc_unit(x)
    if x.to_i>0
      if x.to_i<=10
        return self.proc_single_dig(x.to_i)
      else
        if x.to_i<=20
        return self.proc_double_dig(x.to_i)
        else
        return (self.proc_double_dig((x.to_i/10)*10)) + (self.proc_single_dig(x.to_i%10))
        end
     end
    end
  return ""
  end

  def proc_double_dig(z)
    if z==0
      return ""
    else
      return @h[z]
    end
 end

  def proc_single_dig(y)
    if y==0
      return ""
    else
      return @h[y]
    end
  end
protected :def_calc, :proc_unit, :proc_double_dig,
  :proc_single_dig

end

puts 453645445.english_word

#FourtyFive Crore, Thirtysix Lakh, FourtyFive Thousand,Four Hundred,FourtyFive

参考文献摘自：https://raveendran.wordpress.com/2009/05/29/ruby-convert-number-to-english-word/

我希望它可以帮助你：）

Answer 2

这是我找到的解决方案。我希望这会对别人有所帮助。

<extension point="org.eclipse.jdt.ui.cleanUps">
    <cleanUp id="eclipsecs.saveaction" class="eclipsecs.saveaction.CheckFileCleanUp" />
</extension>