当输入输入 y 时如何循环这个答案

【问题标题】：Making a program than would ask the user for an integer then will asked if they wish to loop or quit当输入输入 y 时如何循环这个
【发布时间】：2022-01-20 11:34:45
【问题描述】：

def interact():
    while True: 
        try:
            num = int(input("Please input an integer: "))
            if (num % 2) == 0:
                print ("{0} is even".format(num))
            else:
                print("{0} is odd".format(num))
            num_two = int(input('Do you want to play again n/Y:'))
       
        except:
            if num_input == "y":
                continue
        finally:
            print("Goodbye")
        print(num_two)

【问题讨论】：

num_two 根本不会是一个数字。

标签： python if-statement try-catch except finally

【解决方案1】：

在再见后使用休息声明：

def interact():
    while True: 
        try:
            num = int(input("Please input an integer: "))
            if (num % 2) == 0:
                print ("{0} is even".format(num))
            else:
                print("{0} is odd".format(num))
            num_two = int(input('Do you want to play again n/Y:'))
       
        except:
            if num_input == "y":
                continue
        finally:
            print("Goodbye")
            break
        print(num_two)

【讨论】：

【解决方案2】：

重播提示要求输入一个字符串，而不是数字。不要试图把它当作一个数字。将try 语句限制为测试数字输入：如果得到ValueError，请立即重新启动循环，而不是尝试测试它是偶数还是奇数。

当您获得再次播放的响应时，如果响应不是y，请使用break 退出循环。

循环退出后打印“再见”。

def interact():
    while True:
        try:
            num = int(input("Please input an integer: "))
        except ValueError:
            continue  # Ask for a number again

        if num % 2 == 0:
            print("{0} is even".format(num))
        else:
            print("{0} is odd".format(num))

        response = input("Play again? (n/Y) ")
        if response != "y":
            break
    print("Good bye")

【讨论】：

非常感谢。我只是 python 的新手。感谢您提供新信息。它会很有帮助

【解决方案3】：

我不完全明白你在追求什么，但也许可以试试这样：

def interact():
    while True: 
        try:
            num = int(input("Please input an integer: "))
            if (num % 2) == 0:
                print ("{0} is even".format(num))
            else:
                print("{0} is odd".format(num))
        except:
            continue
        if input("Would you like to play again? Y/N: ").lower() == "y":
            continue
        else:
            print("goodbye")
            break

我相信这会产生你所追求的效果。

【讨论】：

【解决方案4】：

我认为代码应该是这样的：

def interact():
  try:
      while True:
          num = int(input("Please input an integer: "))
          if (num % 2) == 0:
              print ("{0} is even".format(num))
          else:
              print("{0} is odd".format(num))
          num_two = str(input('Do you want to play again n/Y:'))

          if num_two != "y":
              raise
  except:
      print("Goodbye")

【讨论】：

【解决方案5】：

更紧凑的方法：

def interact():
    want_continue = "Y"
    while want_continue == "Y":
        try:
            num = int(input("Please input an integer: "))
            print ("{0} is {1}".format(num, "even" if num % 2 == 0 else "odd"))
            want_continue = input('Do you want to play again (n/Y):').upper()
        except ValueError:
            print("Please enter a number")
    print("Goodbye")

【讨论】：