Question

我能够倒数到97瓶啤酒，但我无法循环计数到1.是否有可能用我写的东西创建一个循环？这是我到目前为止所拥有的。

all_beers = (99).to_s
one_less = ((all_beers).to_i - 1).to_s
puts '' +
  all_beers + ' bottles of beer on the wall, ' +
  all_beers + ' bottles of beer. You take one down you pass it around ' +
  one_less + ', beers on the wall!'

all_beers = one_less
one_less = ((all_beers).to_i - 1).to_s
puts '' +
  all_beers + ' bottles of beer on the wall, ' +
  all_beers + ' bottles of beer. You take one down you pass it around ' +
  one_less + ', beers on the wall!'

Answer 1

使用downto：

它将从您想要的数字循环到您想要的数字。

99.downto(1).each do |s|
  all_beers = s
  one_less = s - 1
  puts '' +
      all_beers.to_s + ' bottles of beer on the wall, ' +
      all_beers.to_s + ' bottles of beer. You take one down you pass it around ' +
      one_less.to_s + ', beers on the wall!'
end

Answer 2

是的，这当然是可能的。这取自99 Bottles of Beer project：

# 
# Rubeer.rb
# by Eric Budd, Jan. 2008
#
# Demonstrates adding functionality to a built-in class, optional method parameters, inline 
#   conditionals, string replacement, alcohol aversion, and excessively fancy use of hashes.
#
# This borrows the hash from Daniel Straight's excellent implementation for the "wordalize" method
#

class Integer
  NUMBER_WORDS = { 0 => "no", 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"}

  def wordalize
    raise "Invalid number to wordalize - should be in the range (0..99)" unless (0..99) === self
    return NUMBER_WORDS[self] if self < 20
    wordalized = NUMBER_WORDS[self - (self % 10)]
    wordalized += '-' + NUMBER_WORDS[self % 10] unless (self % 10) == 0
    return wordalized
  end

  def bottles
    raise "Invalid number of bottles - should be in the range (0..99)" unless (0..99) === self
    how_many_bottles = self.wordalize + ' bottle'
    how_many_bottles += 's' unless self == 1
    return how_many_bottles 
  end

  alias :bottle :bottles # for grammar Nazis
end

def sing(number, teetotaller = false)
  beverage = teetotaller ? 'coke' : 'beer'

  puts "#{number.bottles.capitalize} of #{beverage} on the wall, #{number.bottles} of #{beverage}."
  if number != 0
    puts "Take one down, pass it around, #{(number - 1).bottles} of #{beverage} on the wall.\n\n"
  else
    puts "Go to the store and buy some more, 99 bottles of #{beverage} on the wall."
  end
end

99.downto(0) { |number| sing(number) }
# Uncomment the following for the alternative teetotaller version
# 99.downto(0) { |number| sing(number, true) }

上传了多个Ruby版本，但默认版本对于初学者来说太复杂了。但是，这个非常好，你应该能够理解发生了什么。

应该回答你问题的一点是99.downto(0) { |number| ... }。这是一个循环，它将重复大括号内的任何内容（在本例中为sing(number)）一百次，number从99转到0。

另请注意，将数字作为字符串（(99).to_s）携带并在需要时将其转换回整数是低效的（并且难以辨认）;相反，让它始终是一个整数，并在你需要它作为字符串之前将其转换为字符串，当你显示时（或者让字符串连接/插值自动为你做，就像在这段代码中一样）。 p>

虽然Ruby确实同时具有for和while个循环，但它们很少使用（while）或从不使用for。相反，Rubyists通常依赖于迭代器和枚举器。 Integer#downto等其他功能是Integer#upto，Integer#times以及最令人敬畏的Enumerable混合中的所有内容。

Answer 3

简短的回答是肯定的，你可以使用你编写的内容进行循环，并且有许多循环使用ruby的方法。然而，由于这似乎是关于学习编程，考虑到你没有使用任何控制结构或字符串插值，更不用说没有多大意义的演员，我建议Why's Poignant Guide to Ruby来在使用ruby时学习编程的概念。

除了downto之外，你还可以做点什么：

(1..99).reverse_each do |number|
  bottle = number == 1 ? 'bottle' : 'bottles'
  verse = "#{number} #{bottle} of beer on the wall, #{number} #{bottle} of beer. "
  verse << "Take one down, pass it around, #{number-1} #{number-1 == 1 ? 'bottle' : 'bottles'} of beer on the wall"
  puts verse
end

Answer 4

您可以用来让生活更轻松的事情：downto()，proc，三元if（ - that - thing - ＆gt;？：）

我很高兴第一次体验这项运动，所以我对在这里提供答案有点犹豫，但事实如此。

它有点先进，但使用了一个＆＃39; proc＆＃39;确保你复制瓶子和＃34;正确是一个很好的，干净的方式来完成它。

＆＃39; DOWNTO（）＆＃39;也是一种很棒的方式来迭代这99瓶，因为它让你觉得你在阅读英语而不是代码。

num_at_start = 99 # You may change this number.

num_bottles = proc { |n| "#{n} bottle#{ n == 1 ? '' : 's'}" }

num_at_start.downto(1) do |num| 
  print  "#{num_bottles.call(num)} of beer on the wall, " +
         "#{num_bottles.call(num)} of beer!\n" +
         "You take one down, pass it around, "
  unless num == 1
    puts "#{num_bottles.call(num - 1)} of beer on the wall!"
  else
    puts "No more bottles of beer on the wall!"
  end
end

来源：学习Chris Pine的第2版程序（我改变了一些事情）

Answer 5

这是我在这个问题上找到的解决方法：

beer = 99
    while beer > 0

    puts beer.to_s + " bottles of beer on the wall. " + beer.to_s +
    " bottles of beer."

     (beer -= 1).to_s

    puts "Take one down, pass it around. " + beer.to_s +
    " bottles of beer on the wall."
end

麻烦用99瓶啤酒循环串

5 个答案: