Question

错误：

Traceback (most recent call last):
  File "C:/Python/CurrencyCoverter/currencyconverter.py", line 16, in <module>
    if userChoice == "1":
NameError: name 'userChoice' is not defined

如果我尝试运行我的货币转换器脚本，这里是脚本（当前未完成）：

def currencyConvert():

    userChoice = input("What do you want to convert? \n1.)USD > UK \n2.)USD > UK \n")


if userChoice == "1":    
    userUSD = imput("ENTERAMOUNT")

    UK = userUSD * 0.62
    print ("USD", userUSD, "= ", UK, "UK")



elif userChoice == "2":
    print ("Choice = 2")

else:
     print ("Error, Please Choose Either Option 1 or 2")

Answer 1

问题是您正在尝试访问userChoice，currencyConvert仅在currencyConvert范围内，在函数外部。

要解决此问题，请userChoice返回userChoice = currencyConvert()，然后按以下方式访问：

def currencyConvert():

    userChoice = input("What do you want to convert? \n1.)USD > UK \n2.)USD > UK \n")

    # Return userChoice
    return userChoice

# Access userChoice (the return value of currencyConvert)
userChoice = currencyConvert()

if userChoice == "1":    
    userUSD = imput("ENTERAMOUNT")

    UK = userUSD * 0.62
    print ("USD", userUSD, "= ", UK, "UK")

elif userChoice == "2":
    print ("Choice = 2")

else:
    print ("Error, Please Choose Either Option 1 or 2")

换句话说，您的代码应如下所示：

{{1}}

Answer 2

首先，我希望缩进在这里搞砸了，而不是在你的实际剧本中;否则，这应该是你的第一要务。

我认为你误解了功能的重点。您正在定义此函数以获取输入，然后将其丢弃（因为它不会被返回）。此外，你永远不会打电话给这个功能。

如果我是你，因为该函数基本上是一行代码，我会完全删除该函数。

此外，else块的内容让我相信脚本的整体形式已被破坏。我会做类似以下的事情：

# I kept the function in this example because it is used twice. In your example, it was only used once, which is why I recommended removing it.
def getChoice():
    return input("What do you want to convert? \n1.)USD > UK \n2.)USD > UK \n")
userChoice = getChoice()
while userChoice != "1" and userChoice != "2": # better yet, you could have a list of valid responses or even use a dictionary of response : callback
    userChoice = getChoice()
# Process input here

名称未定义错误Python

2 个答案: