Question

我正在尝试编写一个允许人们输入句子的代码，之后将检查句子的元音和审查，它还应该返回句子中发现的元音总数。这是我到目前为止所做的，但是不是一次运行输入，而是多次输入，然后代码出现以下错误：TypeError: 'NoneType' object is not iterable。

def vowel(func):
  def wrapper():
    vowels = "aeiouAEIOU"
    vowel_count = 0
    for items in func():
        if items in vowels:
            vowel_count += 1
            print("Vowel found: " + items)
    print("Total amount of vowels", vowel_count)
    return wrapper


def censorship(func):
    def wrapper():
        censorship_list = ["Word", "Word1", "Word2", "Word3"]
        for words in censorship_list:
            if words in func():
                print("You are not allowed to use those word(s) in set order: ", words)
    return wrapper


@vowel
@censorship
def sentence():
    sent = input("Input your sentence: ")
    return sent

sentence()

Answer 1

有两个问题。首先，wrapper函数都没有返回值。因此，当sentence被调用时，wrapper中的第一个censorship将不会返回在vowel中使用的必要值。其次，func中的vowel调用将返回前一个装饰器返回的所有值（在censorship中）;但是，即使wrapper中的censorship返回了一个值，它也必须包含原始字符串输入的副本：

def vowel(func):
  def wrapper():
    vowels = "aeiouAEIOU"
    vowel_count = 0
    start_val = func()
    for items in start_val:
      if items in vowels:
        vowel_count += 1
    return start_val, vowel_count
  return wrapper

def censorship(func):
  def wrapper():
     censorship_list = ["Word", "Word1", "Word2", "Word3"]
     string = func()
     for words in censorship_list:
       if words in string:
         raise ValueError("You are not allowed to use those word(s) in set order: ")
     return string
  return wrapper

@vowel
@censorship
def sentence():
  sent = input("Input your sentence: ")
  return sent

使用装饰器获得意外错误

1 个答案: