英文:
How do you detect specifically an integer input but also check to see if the input to something is blank?
问题
我在想是否有可能让输入变量寻找一个整数输入，同时还能够检测用户是否没有输入。我想制作一个简单的应用程序，用户可以从一个编号列表中选择一个选项，如果用户将输入区域留空，希望程序默认选择第一个选项。
我目前有一个尝试循环，不断要求用户输入整数，写成这样：
def inputNum():
while True:
try:
userInput = int(input())
except ValueError:
print("输入不是整数");
continue
except EOFError:
return ""
break
else:
return userInput
break
selection = inputNum()
我有一种感觉这其中的部分显然是错误的，但我希望每个人都能理解大意并提供有用的反馈和答案。
英文:
I was wondering if it was possible to have an input variable look for an integer input but also be able to detect if the user leaves no input. I was wanting to make a simple app where the user chooses an option out of a numbered list and want the program to default to the first option if the user leaves the input area blank.
I currently have a try loop that continually asks the user for an integer as an input written like this:
def inputNum():
while True:
try:
userInput = int(input())
except ValueError:
print("Input is not an integer");
continue
except EOFError:
return ""
break
else:
return userInput
break
selection = inputNum()
I have a feeling part of this is blatantly wrong, but I hope everyone gets the gist of things and can help provide helpful feedback and answers.
答案1
得分: 0
不要在请求输入时将其转换为整数。将您的输入请求分为两个步骤：
def inputNum():
while True:
userInput = input().strip()
if not userInput:
print('输入为空')
continue
try:
userInput = int(userInput)
except ValueError:
print("输入不是整数");
continue
return userInput
selection = inputNum()
示例：
输入为空
abc
输入不是整数
123
输出：123
英文:
Don't convert to integer while requesting input. Split your input request into 2 steps:
def inputNum():
while True:
userInput = input().strip()
if not userInput:
print('Input is empty')
continue
try:
userInput = int(userInput)
except ValueError:
print("Input is not an integer");
continue
return userInput
selection = inputNum()
Example:
Input is empty
abc
Input is not an integer
123
Output: 123
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