Question

您好，这是我正在处理的代码的一部分。我想知道是否有办法检测输入的输入是否为字母并给出响应。我需要减去时间，所以我将它转换为int，但如果除了int之外的任何东西都是类型，如果给我一个错误。

def Continue():
    time = int(input("What time is it?: "))
    if time > 12:
        print ("I'm sorry, just the hour, I don't need anymore than that.")
        Continue()
    else:
        print ("It's %d" % time + "?!")
        time = time - 1
        print ("I'm late!\nI'm sorry I have to go!\nI'm sure your leg is fine just walk it off!")
        print ("I was suppose to be there at %d" % time, "I'm an hour late!")

Answer 1

这样做你想要的：

def Continue():
    try:
        time = int(input("What time is it?: "))
    except ValueError:
        print('Invalid input.')
        print('Please enter a number between 1 and 12.')
        Continue()
    if time > 12:
        print("I'm sorry, just the hour, I don't need anymore than that.")
        Continue()
    else:
        print("It's %d" % time + "?!")
        time = time - 1
        print ("I'm late!\nI'm sorry I have to go!\nI'm sure your leg is fine just walk it off!")
        print ("I was suppose to be there at %d" % time, "I'm an hour late!")

您可以使用try - except声明。如果转换为整数失败，Python将引发ValueError。在这种情况下，您使用except捕获异常，告诉用户输入错误，然后再次调用您的函数。

Answer 2

解决此问题的一种方法是不立即将输入转换为int，而是创建一个if else语句，用于检查输入是否为数字。如果是数字，则可以将变量转换为int。

检测输入是否是python 3.5中的字母？

2 个答案: