Python返回那个数字输入不正确，但是是正确的

Question

我制作了一个 python 程序，可以在随机数或字母之间进行选择，然后用户尝试猜测它，但是当程序选择一个数字并正确猜出确切数字时，它returns 用户输入的数字不正确

这是我的代码：

import math
import random
import string


count = 0
numberroll = random.randint(1,10)
stringroll = random.choice(string.ascii_letters)

chooseroll = random.choice([numberroll, stringroll.lower()])

# print(chooseroll)

count = 0
if chooseroll == numberroll:
    usernumberguess = input("Your guess: ")
    if usernumberguess.isdigit():
        print("It is a number!")
        count += 1
    else:
        print("That is not a number...")
    if usernumberguess.isdigit() == numberroll:
        print("Correct number!")
        count += 2
    else:
        print("Better luck next time!")

if chooseroll == stringroll.lower():
    userstringguess = input("Your guess: ")
    if userstringguess.isalpha():
        print("It is a letter!")
        count += 1
    else:
        print("That is not a letter...")
    if userstringguess.lower() == stringroll.lower():
        print("Correct letter!")
        count += 2
    else:
        print("Incorrect! Better luck next time!")

# print(count)

有什么办法可以解决这个问题吗？提前致谢！

Answer 1

这部分if usernumberguess.isdigit() == numberroll:可能有问题：isdigit() returns对错，你看看是不是numberroll。您可以在猜测是数字的情况下检查猜测是否正确。您还需要将字符串形式的输入转换为 int.

if chooseroll == numberroll:
    usernumberguess = input("Your guess: ")
    if usernumberguess.isdigit():
        print("It is a number!")
        usernumberguess = int(usernumberguess)
        count += 1
        if usernumberguess == numberroll:
            print("Correct number!")
            count += 2
        else:
            print("Better luck next time!")
    else:
        print("That is not a number...")

我会添加一些更多的代码用于调试目的，这样您就可以看到哪里出了问题：

import random
random.seed(0)
count = 0
numberroll = random.randint(0, 9)
print(f"the user has to guess: {numberroll}")

usernumberguess = input("your guess: ")
print(f"your guess is {usernumberguess}")

if usernumberguess.isdigit():
    print("your guess is a number")
    usernumberguess = int(usernumberguess)
    count += 1
    if usernumberguess == numberroll:
        print("your guess is correct")
        count += 2
    else:
        print("your guess is not correct")
else:
    print("your guess is not a number")

Python返回那个数字输入不正确，但是是正确的

Python returning that number input is incorrect, but it is correct

python

string

random

numbers

choice