Python初学者编写四则运算计算器时始终触发Unknown Entry错误如何解决

问题原因 #

• 第一个逻辑错误在输入合法性判断的条件运算符使用错误：你用了or（或）来拼接多个!=判断，无论输入什么值，必然会满足其中一个不等条件，导致判断始终为真，一直输出错误提示。正确应该用and（且），即输入同时不等于1/2/3/4的时候才判定为非法输入。
• 第二个错误是非法输入分支下的UnKnownEntry == True是比较运算，不是赋值运算，没有真正修改UnKnownEntry变量的值，导致循环逻辑异常。

修正后的完整代码 #

OperationSelected = False
UnKnownEntry = False

while OperationSelected == False or UnKnownEntry == True:
    try:
        operation = input("Select operation to perform:\r\n1>PLUS\r\n2>MINU\r\n3>MULTIP\r\n4>DIVIDE\r\n>>> ")

        if operation != "1" and operation != "2" and operation != "3" and operation != "4":
            print("--------------------------")
            print("Unknown entry. ( " + operation + " )")
            print("--------------------------")
            UnKnownEntry = True
        else: 
            print("TEST")
            OperationSelected = True
            UnKnownEntry = False

    except Exception as e:
        print(e)
        break

while OperationSelected == True and UnKnownEntry == False:
    try:    
        num1 = input("Enter first number: ")
        num2 = input("Enter second number: ")
        
        if operation == "1":
            print("--------------------------------------")
            print("The Solution is: " + str(float(num1) + float(num2)))
            print("--------------------------------------")
        elif operation == "2":
                print("--------------------------------------")
                print("The Solution is: " + str(float(num1) - float(num2)))
                print("--------------------------------------")
        elif operation == "3":
            print("--------------------------------------")
            print("The Solution is: " + str(float(num1) * float(num2)))
            print("--------------------------------------")
        elif operation == "4":
            print("--------------------------------------")
            print("The Solution is: " + str(float(num1) / float(num2)))
            print("--------------------------------------")
    except Exception as e:
        print(e)
        break

可选优化 #

合法性判断可以简化为成员运算，可读性更高：

if operation not in ["1", "2", "3", "4"]:

内容的提问来源于stack exchange，提问作者Mr.RobotFAN