Question

我正在尝试从用户处获取输入，如果该输入为空，则应该看到带有（请输入数字）的消息

我尝试使用if语句并将其设置为等于None。

def get_player_numbers():
    number_csv = input("Enter your 6 numbers, separated by commas: ")
    if number_csv == None:
        print("Enter something")
        get_player_numbers()
    else:
        number_list = number_csv.split(",")
        integer_set = {int(number) for number in number_list}
        return integer_set

我应该能够看到（“请输入数字，输入不能为空”）

Answer 1

input()始终返回一个字符串。空的输入返回一个空字符串：

if number_csv == '':
        print("Please enter digits , input can't be blank")

由于隐含虚假的空字符串，可以将其简化为：

if not number_csv:
        print("Please enter digits , input can't be blank")

Answer 2

我认为这不会编译，因为您将收到文件结束错误。您应该使用raw_input（）而不是input（）。区别在于input（）返回python表达式类型的对象，而raw_input返回一个字符串。在这种情况下，您将要比较字符串。您的新代码将是...

def get_player_numbers():
    number_csv = raw_input("Enter your 6 numbers, separated by commas: ")
    if number_csv == "":
        print("Enter something")
        return get_player_numbers()
   else:
        number_list = number_csv.split(",")
        integer_set = {int(number) for number in number_list}
        return integer_set

注意：现在，您的if语句会将输入值与“”而不是“ None”进行比较。

编辑：感谢您的评论！如果用户输入空格，我通过添加return语句来修复递归调用。这需要发生，或者当用户最终输入内容时，输出将为“无”。

Answer 3

我建议您使用正则表达式（test）来测试输入：

import re


def get_player_numbers():
    integer_set = None
    while not integer_set:
        number_csv = input("Enter your 6 numbers, separated by commas: ")
        if not re.match("([0-9]+,){5}[0-9]+", number_csv):
            print("Invalid input.")
        else:
            number_list = number_csv.split(",")
            integer_set = {int(number) for number in number_list}
    return integer_set


res = get_player_numbers()

print(res)

输出：

Enter your 6 numbers, separated by commas: 
Invalid input.
Enter your 6 numbers, separated by commas: 1,2
Invalid input.
Enter your 6 numbers, separated by commas: some-word
Invalid input.
Enter your 6 numbers, separated by commas: 1,2,3,4,5,6

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3 个答案: