Question

我在弄清楚为什么在包含Docstring时我的代码失败的问题，特别是在函数select_level():上如果我删除文档字符串并注释掉，它会很好，但是如果我包含Docstring，则在终端上会出现以下错误：

（NameError：未定义名称“ select_level”）

def select_level():
    """Defines how a player selects a difficulty, selects questions 
       and answers depending on user input,outputs selections.
    """
    print ("Ready Player One! Select a level.")
    level_name = raw_input("Type in easy, medium or hard\n").lower()
    if level_name=="easy":
        level(easy_level, blanks, easy_answers)
    elif level_name=="medium":
        level(medium_level, blanks, medium_answers)
    elif level_name=="hard":
        level(hard_level, blanks, hard_answers)
    else:
        print ("Please select easy, medium or hard")
    print select_level()

Answer 1

我有一些建议：

确保select_level（）被正确调用并且确实已定义。

我问是因为它说

(The error says (NameError: name 'select_level' is not defined).

还，您是否尝试过在最后一行添加括号？

print(select_level())

Answer 2

您的问题很可能是缩进。您的代码中有多个功能吗？如果是这样，请确保select_level函数未嵌套在另一个函数中。提供您的代码时，我无法重现您的错误。但是当我将您的函数嵌套在这样的另一个函数中时：

def my_other_function():
  print "This is my other function"

  def select_level():
    """Defines how a player selects a diffuclity, selects questions 
      and answers depending on user input,outputs selections.
    """
  print ("Ready Player One! Select a level.")
  level_name = raw_input("Type in easy, medium or hard\n").lower()
  if level_name=="easy":
      level(easy_level, blanks, easy_answers)
  elif level_name=="medium":
      level(medium_level, blanks, medium_answers)
  elif level_name=="hard":
      level(hard_level, blanks, hard_answers)
  else:
      print ("Please select easy, medium or hard")

然后尝试调用select_level函数，该函数会因您遇到的相同错误而中断。看到第二个函数比第一个函数缩进更多吗？这就是导致错误的原因。

由于该函数嵌套在另一个函数中，因此超出了范围。有关python范围的更多信息，give this page a read。

此外，您提供的代码示例还有另外两个问题。

删除代码最后一行的缩进。所以看起来像这样：

def select_level():
  """Defines how a player selects a diffuclity, selects questions 
  and answers depending on user input,outputs selections.
  """
  print ("Ready Player One! Select a level.")
  level_name = raw_input("Type in easy, medium or hard\n").lower()
  if level_name=="easy":
    level(easy_level, blanks, easy_answers)
  elif level_name=="medium":
    level(medium_level, blanks, medium_answers)
  elif level_name=="hard":
    level(hard_level, blanks, hard_answers)
  else:
    print ("Please select easy, medium or hard")
print select_level()

在函数上使用print语句时，它将尝试打印返回值。您的函数没有返回值，因此根本不需要print语句。删除它，使您的代码如下所示：

def select_level():
  """Defines how a player selects a diffuclity, selects questions 
  and answers depending on user input,outputs selections.
  """
  print ("Ready Player One! Select a level.")
  level_name = raw_input("Type in easy, medium or hard\n").lower()
  if level_name=="easy":
    level(easy_level, blanks, easy_answers)
  elif level_name=="medium":
    level(medium_level, blanks, medium_answers)
  elif level_name=="hard":
    level(hard_level, blanks, hard_answers)
  else:
    print ("Please select easy, medium or hard")
select_level()

docstring导致函数未定义

2 个答案: