如何从Python中该函数所在的行继续代码？

Question

我的文字游戏中有一个nameType()函数，并且使用了两次：

1. 创建一个stats类（或其他任何类），以便程序可以保存它们，以及
2. 游戏中用于某些变量更改的功能。

这是此职位针对的当前代码（可能会更改）：

def TitleScreen():
blankHuge()
anar()
print('The Game of Hierarchy and Anarchy\nBy Roach\nText Edition\n▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬\nType "new" to start a new game.\nTo load a game, type "load".\nTo exit, type "exit".')
StartGame = input()
if StartGame == 'new':
    print('► Crew Member: Hello, fine man. I can see you are on an adventure to the land of Salbury.\nBefore you disembark the boat, please enter your name.')
    for filename in file:
        os.unlink(filename)
    nameType()
    global PlayerIG                   
    PlayerIG = Player(CharName)
    print('► Crew Member: Thanks for travelling with us, and have a great time,',PlayerIG.name,'.')
    saveGame()
    MainMenu()
# more stuff that aren't related down here...

这是nameType()函数：

def nameType():
print('▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬')
CharName = input()
if CharName != '':
    if len(CharName) > 20:
        print('► The name is too long!')
        nameType()
    else:
        print('► Are you sure this is what you want as a name?\nIf so, type "yes".\nOtherwise, type "no".')
        confirmName = input()

        if confirmName == 'yes': # If this is true, then jump to next codeline from where the nameType() func was placed...
            pause() # Doesn't work

        if confirmName == 'no':
            nameType()
        else:
            print('► Invalid Input!')
            nameType()

这是应该从原处继续执行代码的函数，但是它不起作用：

def pause():
pause = input('► Press Enter to continue.')
if pause != '':
    pause()

有什么办法可以继续执行我的代码？

示例（很可能无法正常工作）：

def TitleScreen():
    ...
    nameType() # Someway, jump to next line after code in nameType() has been executed...
    global PlayerIG
    PlayerIG = Player(CharName) 
    ...
    saveGame()
    MainMenu() 

def management():
    ...
    opt = input()
    if opt = 'name':
        nameType() # Someway, jump to next line after code in nameType() has been executed...
        PlayerIG.nameWait = 12
        PlayerIG.heat = 0
    ...

注意：我对这将如何工作有一个想法，但是我不确定它是否会工作。 它是：

# outside of nameType()
if confirmName == 'yes':
    ...

Answer 1

我的想法奏效了， if在nameType()函数外部使用， if 确实可行。 我使用了global （我知道这是不好的编码，但是不管用什么），现在该函数起作用了。

def nameType():
global confirmName
global CharName
print('▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬')
CharName = input()
if CharName != '':
    if len(CharName) > 20:
        print('► The name is too long!')
        nameType()
    else:
        print('► Are you sure this is what you want as a name?\nIf so, type "yes".\nOtherwise, type "no".')
        confirmName = input()
# Items down here for the confirmName input removed

# This is the system for the 'jump':
if confirmName == 'blah1':
    # add stuff here...    
    if confirmName == 'no':
        # add stuff here...
    else:
        # add stuff here...