python - 值错误 : invalid literal for int() with base 10 -- how to guard against invalid user input?

Question

我制作了一个程序，用户可以在其中输入一个数字，该程序会计算到该数字并显示花费了多少时间。但是，每当我输入字母或小数(即 0.5)时，都会出现错误。这是完整的错误消息:

Traceback (most recent call last):
  File "C:\Documents and Settings\Username\Desktop\test6.py", line 5, in <module>
    z = int(z)
ValueError: invalid literal for int() with base 10: 'df'

我该怎么做才能解决这个问题？

完整代码如下:

import time
x = 0
print("This program will count up to a number you choose.")
z = input("Enter a number.\n")
z = int(z)
start_time = time.time()
while x < z:
    x = x + 1
    print(x)
end_time = time.time()
diff = end_time - start_time
print("That took",(diff),"seconds.")

请帮忙!

Answer 1

好吧，确实有一种方法可以“解决”这个问题，它的行为符合预期——您不能将字母大小写转换为 int，这实际上没有意义。你最好的选择(这是一种 pythonic 的做事方式)是简单地编写一个带有 try...except block 的函数:

def get_user_number():
    i = input("Enter a number.\n")
    try:
        # This will return the equivalent of calling 'int' directly, but it
        # will also allow for floats.
        return int(float(i)) 
    except ValueError:
        #Tell the user that something went wrong
        print("I didn't recognize {0} as a number".format(i))
        #recursion until you get a real number
        return get_user_number()

然后您将替换这些行:

z = input("Enter a number.\n")
z = int(z)

与

z = get_user_number()