关于计算器程序的问题在Python 3中重复的num1和num2 for else

Question

我决定编制一个计算器，但我的问题是，如果我做了选择并输入6，它仍然会向我询问 num1 和在说＆＃34;输入无效之前，请先检查选择操作并再试一次！＆＃34;。我想要的是它直接说＆＃34;输入无效，请检查选择操作并再试一次！＆＃34;无需请求 num1 和 num2 。我希望它清楚。

def calculator():
def add(x, y):
    return x + y

def sub(x, y):
    return x - y

def mul(x, y):
    return x * y

def div(x, y):
    return x / y

print(""" -- Select Operation --
1. Addition
2. Subtraction
3. Multiplication
4. Division
""")

choice = input("Enter 1 or 2 or 3 or 4 from the Select Operation >> ")
print("\n")
print("Calculating...")
num1 = int(input("Enter First number >> "))
num2 = int(input("Enter Second number >> "))

if choice == '1':
    print(num1, "+", num2, "=", add(num1, num2))
elif choice == '2':
    print(num1, "-", num2, "=", add(num1, num2))
elif choice == '3':
    print(num1, "-", num2, "=", add(num1, num2))
elif choice == '4':
    print(num1, "-", num2, "=", add(num1, num2))
else:
    print("Invalid Input, PLease check the *Select Operation* and Try Again!")

计算器（）

Answer 1

更正了您的代码，请阅读python语法

def add(x, y):
    return x + y

def sub(x, y):
    return x - y

def mul(x, y):
    return x * y

def div(x, y):
    return x / y

print(""" -- Select Operation --
1. Addition
2. Subtraction
3. Multiplication
4. Division
""")

choice = input("Enter 1 or 2 or 3 or 4 from the Select Operation >> ")
while(choice not in ('1','2','3','4')):
  choice = input("Invalid Input Please Enter 1 or 2 or 3 or 4 from the Select Operation >> ")
print("\n")
print("Calculating...")
num1 = int(input("Enter First number >> "))
num2 = int(input("Enter Second number >> "))

if choice == '1':
    print(num1, "+", num2, "=", add(num1, num2))
elif choice == '2':
    print(num1, "-", num2, "=", sub(num1, num2))
elif choice == '3':
    print(num1, "-", num2, "=", mul(num1, num2))
elif choice == '4':
    print(num1, "-", num2, "=", div(num1, num2))