Python猜测游戏拒绝无效的用户输入答案

【问题标题】：Python Guessing Game Reject Invalid User InputPython猜测游戏拒绝无效的用户输入
【发布时间】：2018-11-05 01:07:26
【问题描述】：

我将参加我的第一个 Python 课程，并且与大多数 Python 课程一样，最后一项任务是创建一个 1-100 的猜谜游戏，以跟踪 VALID 尝试的次数。我无法获得（或在 stackoverflow 上找到）的元素是如何拒绝无效的用户输入。用户输入必须是整数，介于 1 和 100 之间的正数。我可以让系统拒绝除 0 和

我唯一能想到的就是告诉我不能让运算符比较字符串和整数。我一直想使用诸如guess > 0 和/或guess

#  Generate random number
import random
x = random.randint(1,100)

#  Prompt user for input
print("I'm thinking of a number from 1 to 100")
counter = 0

while True:
    guess = input("Try to guess my number: ")

    #  Check if input is a positive integer and is not 0 or >=101
    #  this line doesn't actually stop it from being a valid guess and 
    #  counting against the number of tries.
    if guess == "0":  
        print(guess, "is not a valid guess")
    if guess.isdigit() == False: 
        print(guess, "is not a valid guess")
    else: 
        counter += 1
        guess = int(guess)

        #  Begin playing
        if guess > x:
            print(guess, "is too high.")
        elif guess < x:
            print(guess, "is too low.")
        else:
            print(guess, "is correct! You guessed my number in", counter, "tries!")

【问题讨论】：

查看 Try 以及如果 try 失败如何处理。用户将传递一个单词，您尝试将其转换为 int，如果失败，则给出输入无效的输出。
Asking the user for input until they give a valid response的可能重复
int(guess) 应该是第一件事。如果失败，则它不是整数。然后你只需要检查它是小于 1 还是大于 100，这很容易，因为它已经被转换为整数了。
与其单独考虑猜测无效的所有情况，不如尝试为猜测是否有效创建一个测试。

标签： python

【解决方案1】：

import random
x = random.randint(1,100)

#  Prompt user for input
print("I'm thinking of a number from 1 to 100")
counter = 0

while True:
    guess = input("Try to guess my number: ")
    try:
        guess = int(guess)
        if(100 > guess > 0):
            counter += 1
            guess = int(guess)
            #  Begin playing
            if guess > x:
                print(guess, "is too high.")
            elif guess < x:
                print(guess, "is too low.")
            else:
                print(guess, "is correct! You guessed my number in", counter, "tries!")
                break
        else:
            print("Number not in range between 0 to 100")
    except:
        print("Invalid input")

【讨论】：

【解决方案2】：

#  Generate random number
import random
x = random.randint(1,100)

#  Prompt user for input
print("I'm thinking of a number from 1 to 100")
counter = 1   


while True:
    try:
        guess = int(input("Try to guess my number: "))
        if guess > 0 and guess < 101:
            print("That's not an option!")
        #  Begin playing
        elif guess == x:
            print(guess, "is correct! You guessed my number in", counter, "tries!")
            break
        elif guess > x:
            print(guess, "is too high.")
        elif guess < x:
            print(guess, "is too low.")
        else:
            counter += 1        
    except:
        print("That's not a valid option!")

【讨论】：

【解决方案3】：

我的导师帮助了我。（我发帖是为了避免需要那个给我评分的人。）这就是我们想出的。我发布它是为了帮助任何未来可能遇到这种特殊拒绝用户输入问题的 Python 学习者。

感谢你们这么快发帖！即使我需要导师的帮助，如果没有你的见解，我会显得更加无能。现在我可以真正享受我的假期周末了。有一个伟大的阵亡将士纪念日周末！！！

import random
x = random.randint(1,100)

print("I'm thinking of a number from 1 to 100.")
counter = 0

while True:
    try:        
        guess = input("Try to guess my number: ")
        guess = int(guess) 
        if guess < 1 or guess > 100:
            raise ValueError()
        counter += 1
        if guess > x:
            print(guess, "is too high.")
        elif guess < x:
            print(guess, "is too low.")
        else:
            print(guess, "is correct! You guessed my number in", counter, "tries!")
            break
    except ValueError:
        print(guess, "is not a valid guess")

【讨论】：