使用内部用户输入作为函数Python中的输入

debugcn 发表于 Dev

小红帽

我目前遇到一些麻烦这在Python 3。我试图找出如何使用存储在用户输入income并pension_contribution在pension()，但什么也没有发生。这些用户输入应该放在哪里才能被认为是输入，以便功能可以工作？

income = input(int("What is your annual income?"))
pension_contribution = input(int("What is your monthly pension contribution?"))

def pension(pension_contribution, income):
    if pension_contribution > ((income//12) * .075):
        print ("Your pension contribution is too high. Monthly pension contributions may not exceed 7.5% of your gross monthly income")
    else:
        print("Your annual pension contribution is R", pension_contribution * 12)

逻辑

您可以通过3种不同的方式执行此操作：

1.只需调用该函数

def pension (pension_contribution, income):

    if pension_contribution > ((income//12) * .075):
        print ("Your pension contribution is too high. Monthly pension contributions may not exceed 7.5% of your gross monthly income")

    else:
        print("Your annual pension contribution is R", pension_contribution * 12)

income = int(input("What is your annual income?"))
pension_contribution = int(input("What is your monthly pension contribution?")) 

pension(pension_contribution, income) #call function

2.将变量放入函数中

def pension ():
    income = int(input("What is your annual income?"))
    pension_contribution = int(input("What is your monthly pension contribution?"))
    if pension_contribution > ((income//12) * .075):
        print ("Your pension contribution is too high. Monthly pension contributions may not exceed 7.5% of your gross monthly income")

    else:
        print("Your annual pension contribution is R", pension_contribution * 12)


pension() #still have to call function

2.全局变量（不推荐）

income = int(input("What is your annual income?"))
pension_contribution = int(input("What is your monthly pension contribution?"))

def pension ():
    global income, pension_contribution
    if pension_contribution > ((income//12) * .075):
        print ("Your pension contribution is too high. Monthly pension contributions may not exceed 7.5% of your gross monthly income")

    else:
        print("Your annual pension contribution is R", pension_contribution * 12)

pension() #still have to call function

请注意，每次必须如何实际“调用”该函数。否则，程序将成功运行，但永远不会打扰进入该功能。

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