简单的Python输入错误 [英] Simple Python input error

问题描述

我正在尝试编写代码来编辑列表并使之成为回文.

I'm trying to write a code to edit a list and make it a palindrome.

一切正常，除了我的输入仍然给我一个错误.当我在get_number_2中输入非整数时，它将崩溃.

Everything is working except my input still gives me one error. When I enter a non-int into get_number_2, it crashes.

def get_number():
    num = raw_input("Please enter number between 100,000 and 1,000,0000: ")
    if not num.isdigit():
        print "---------------------------"
        print "Invalid input: numbers only"
        print "---------------------------"
        my_main()
    else:
        return num

def get_number_2(n):
    num = input("Please confirm the number you have entered: ")
    if num != int(n):
        print "--------------------"
        print "Entries do not match"
        print "--------------------"
        my_main()
    else:
        return num

我将get_number_2的输入用于其余的代码，因为get_number在检查两个数字之间是否不起作用时不起作用.

I use the input from get_number_2 for the rest of the code as get_number doesn't work when I check if its between two numbers.

有什么方法可以验证输入是否为get_number_2中的int，以便摆脱get_number?

Is there any way i can validate if input is an int in get_number_2 so that I can get rid of get_number?

推荐答案

您还应该使用raw_input和int(num):

def get_number_2(n):
    num = raw_input("Please confirm the number you have entered: ")
    if not num.isdigit() or int(num) != n:
        print "--------------------"
        print "Entries do not match"
        print "--------------------"
        my_main()
    else:
        return int(num)

注释:

• 我假设参数n是一个整数，或者要检查此参数，可以将if更改为:if not num.isdigit() or not n.isdigit() or int(num) != int(n).
• 通过使用 isdigit 来检查是否真正转换为int之前的整数.

• I assume that the parameter n is an int, or to check this you could change the if to: if not num.isdigit() or not n.isdigit() or int(num) != int(n).
• By using isdigit we check if it is an integer before really converting it to int.

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