Ruby 不将“downcase”或“split”识别为已定义的方法？ [复制]

Question

问题在于以下代码：

#write your code here
def translate phrase
  phrase = phrase.downcase.split(/ /)
  phrase.collect! do |word|
    word = word.split(//)
    switched = false
    while switched == false
      word.each_index do |letter| 
        if word[letter] == ("a" || "e" || "i" || "o" || "u")
          switched = true
          word = (word[letter..-1] + word[0..(letter-1)]).join + "ay"
        end
      end
    end
  end
  return phrase.join(" ")
end

puts translate("chocolate cream")
#Should return "ocolatechay eamcray"

当我运行它时，Ruby 只返回一个空行。因此，为了解决问题，我将定义加载到 repl 中。 repl 返回以下错误：

NoMethodError: undefined method `downcase' for #<RubyVM::InstructionSequence:0x000000016e8f88>
    from /home/adc/odin-project/web-development-101/21-ruby-tdd/ruby_tdd_project/learn_ruby/04_pig_latin/pig_latin.rb:3:in `translate'
    from /home/adc/.rvm/rubies/ruby-2.3.0/bin/irb:11:in `<main>'

如果我从代码中删除 downcase，我会收到相同的错误消息，只是这次是 split。

这里有什么问题？ （我非常有信心 downcase 和 split 不是问题。）

Answer 1

If 返回一个空白字符串，因为您在 collect! 块内没有返回任何内容。返回word 就可以了：

def translate phrase
  phrase = phrase.downcase.split(/ /)
  phrase.collect! do |word|
    word = word.split(//)
    switched = false
    while switched == false
      word.each_index do |letter|
        if word[letter] == ("a" || "e" || "i" || "o" || "u")
          switched = true
          word = (word[letter..-1] + word[0..(letter-1)]).join + "ay"
        end
      end
    end
    word
  end
  return phrase.join(" ")
end

puts translate("chocolate cream")
#=> atechocolay amcreay

它看起来不像你所期望的那样返回，但它仍然比空白字符串好。

至于你在控制台中出现的奇怪错误信息，似乎是 REPL 特有的（可能是因为方法名translate）。

Answer 2

更新 1（我将如何编写此代码）

def translate phrase
  phrase = phrase.downcase.split(/ /)
  phrase.collect! do |word|
    word = word.split(//)

    word.each_index do |letter| 
      if ["a", "e", "i", "o", "u"].include?(word[letter])
        if letter == 0
          word = word[letter..-1].join + "ay"
        else
          word = (word[letter..-1] + word[0..(letter-1)]).join + "ay"
        end

        break
      end
    end

    word
  end

  return phrase.join(" ")
end

puts translate("chocolate cream")

更新 2（如果我对您的代码进行一些小改动以使其有用）

def translate phrase
  phrase = phrase.downcase.split(/ /)
  phrase.collect! do |word|
    word = word.split(//)
    switched = false
    while switched == false
      word.each_index do |letter| 
        if ["a", "e", "i", "o", "u"].include?(word[letter])
          switched = true
          if letter == 0
            word = word[letter..-1].join + "ay"
          else
            word = (word[letter..-1] + word[0..(letter-1)]).join + "ay"
          end
          break
        end
      end
    end
    word
  end
  return phrase.join(" ")
end

puts translate("chocolate cream")

说明

1. "o"==("a" || "e" || "i" || "o" || "u")

   和

   "o"== "a" || "o" == "e" || "o" == "i" || "o" == "o" || "o" == "u"

   语句不一样。第一个是假的

   （因为("a" || "e" || "i" || "o" || "u")=="a"和"o"!="a"），因为第二个是真的。

2. 你需要内部 if 语句

   if letter == 0
     word = word[letter..-1].join + "ay"
   else
     word = (word[letter..-1] + word[0..(letter-1)]).join + "ay"
   end
   

   因为当单词以任何提到的元音开头时，然后

   word[letter..-1] + word[0..(letter-1)]

   语句将返回整个单词两次，因为字母将等于 0。

3. 最后你需要在while循环之后返回word对象。