Python 嵌套的if语句不起作用

好的，我正在做一个基于文本的冒险，我引入了变量gold。我正在努力使它在一个故事的选择，你有一个选择，拿起黄金。我的语法如下所示：

if choice == "A":
    print("There is no answer. But there is 5 gold on the floor, would you like to pick it up? Type: YES or NO")
yesorno = input()
if yesorno == "YES":
    g = g + 5
    print("You picked them up.",ge,"g")


elif choice == "B":
    print("Someone from a long distance away shouts: 'Shut up",name,"!'. Then the man walks away down what seems like a echoey corridor.")

elif choice == "C":
    print("Nothing happens, you are left to die.")
    sys.exit("You lost")

选择“A”很好，但如果我想选择B或C，我必须键入两次，如下所示：

Either type A, B or C to choose.
B
B
Someone from a long distance away shouts: 'Shut up g !'. Then the man walks away down     what seems like a echoey corridor.

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假设您使用的是python 3，我认为问题在于缩进和未定义的变量

ge

if choice == "A":
    print("There is no answer. But there is 5 gold on the floor, would you like to pick it up? Type: YES or NO")
    yesorno = input()
    if yesorno == "YES":
        g = g + 5
        print("You picked them up.",g,"g")
elif choice == "B":
    print("Someone from a long distance away shouts: 'Shut up",name,"!'. Then the man walks away down what seems like a echoey corridor.")
elif choice == "C":
    print("Nothing happens, you are left to die.")
    sys.exit("You lost")

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那么你得到的错误是什么？给出错误的选择代码B在哪里？你的意思是当结果“B”放在yesorno中时选择B会返回错误？如果有错误，请提供回溯和代码的相关部分（这意味着）你所说的代码行和错误都不在问题中。请完成你的问题！确保缩进正确。看起来

yesorno

属于choice

A

，因此它需要相同的缩进。变量ge的定义更进一步，但这就把它全部整理好了，非常感谢：）