强行插入if/else语句的代码

Question

我正在制作一个多重计算器，我几乎把它弄坏了，问题是，当我请求一个数字时，如果用户给出一个字符串输入，即使我有一个用于else:语句的if，代码也会中断并抛出一个错误。

def Start():
    numberOneList = []
    numberTwoList = []
    multiples = 100000
    iterations = 0
    multiplicity = int(input("How many common multiplicities you would like to find between two numbers: "))

    if multiplicity > 0 and multiplicity < 100001:
        numberOne = int(input("Input the first number: "))
        if numberOne > 0 and numberOne < 100001:
            numberTwo = int(input("Input the second number: "))
            if numberTwo > 0 and numberTwo < 100001:
                for i in range(multiples):
                    mNumberOne = numberOne * i
                    numberOneList.append(mNumberOne)
                    mNumberTwo = numberTwo * i
                    numberTwoList.append(mNumberTwo)
                print("")
                print("Common multiplicities:")
                print("")
                print("Calculating...")
                print("")
                for i in numberOneList:
                    for a in numberTwoList:
                        if a == i:
                            if a != 0:
                                print(numberOne, "x", i / numberOne, "=", i)
                                print(numberTwo, "x", a / numberTwo, "=", a)
                                print("")
                                iterations += 1
                                if iterations == multiplicity:
                                    Again()
                                else:
                                    continue
                            else:
                                continue
                        else:
                            continue
            else:
                print("Invalid answer, restarting")
                Start()
        else:
            print("Invalid answer, restarting")
            Start()
    else:
        print("Invalid answer, restarting")
        Start()

def Again():
    calculateAgain = input("Calculate again? [y/n]: ")
    if calculateAgain == "y":
        Start()
    if calculateAgain == "n":
        quit()
    else:
        Again()

Start()

Answer 1

您得到了一个错误，因为您在用户的输入上断言一个int类型，而不首先检查它是否有效。正如@kindall在comment中所指出的，try/except允许您捕获失败的类型断言并优雅地处理它们。用下面这样的东西包装您的int转换应该可以做到这一点：

try:
    multiplicity = int(input("How many common multiplicities you would like to find between two numbers: "))
except ValueError as e:
    print('Please input a valid number')
    return str(e)

(请注意，您可能只是重复提示，而不是返回错误文本。)

Answer 2

当前代码的一个快速修复方法是创建一个函数，以获取输入或无效值(如果不是int )。

def input_positive_integer(message):
    while(True):
        try:
            value = int(input(message))
            if (value<=0 or value > 100000 ):
                raise ValueError("Not in range")
            break
        except ValueError as e:
            print("Value error!! Try again!")

然后，您可以更改接收整数的所有输入，以使用此函数。

注意，它将保持循环状态，直到该值有效为止。

我真的建议您避免对代码进行递归，尝试使用一个简单的while循环.