开发者

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

开发者 https://www.devze.com 2023-03-14 05:53 出处:网络
I made a program where the user enters a number, and the program would count up to that number and display how much time it took. However, whenever I enter letters or decimals (i.e. 0.5), I would get

I made a program where the user enters a number, and the program would count up to that number and display how much time it took. However, whenever I enter letters or decimals (i.e. 0.5), I would get a error. Here is the full error message:

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'

What can I do to fix this?

Here is the full code:

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_开发者_如何学编程time
print("That took",(diff),"seconds.")

Please help!


Well, there really a way to 'fix' this, it is behaving as expected -- you can't case a letter to an int, that doesn't really make sense. Your best bet (and this is a pythonic way of doing things), is to simply write a function with a try... except block:

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()

You would then replace these lines:

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

with

z = get_user_number()


Try checking

if string.isdigit(z):

And then executing the rest of the code if it is a digit.

Because you count up in intervals of one, staying with int() should be good, as you don't need a decimal.

EDIT: If you'd like to catch an exception instead as wooble suggests below, here's the code for that:

try: 
    int(z)
    do something
except ValueError:
    do something else
0

精彩评论

暂无评论...
验证码 换一张
取 消