在有限尝试游戏中猜我的号码

Question

作为我大学课程的一部分，我正在学习 python 一个任务，我一直在尝试（重新）编写这个猜数字游戏，如果用户在 5 次尝试内未能正确猜出，则终止：

    # Guess My Number Mod 5 tries or bust


import random  

print("\nI'm thinking of a number between 1 and 100.")
print("Try to guess it in five tries or less")

my_number = random.randint(1, 100)
guess = int(input("Go on, Take a guess, I dare ya "))
tries = 1

while guess != my_number:
    if guess > my_number:
        print("Lower...")
    else:
        print("Higher...")
guess = int(input("Go on, Take a guess, I dare ya "))
tries += 1
if tries==5:
        input("You failed to guess the number was it that hard?\n Press any key to exit!)"

print("Well done you guessed correctly!The number was", my_number)
print("And it only took you", tries, "tries!\n")

input("\n\nPress the enter key to exit.")

我认为终止原因不起作用，因为我的 if 语句在 while 循环之外，我无法使其生效。

还有一些无效的语法，因为我累了，看不出来。

如果可能的话，你们能否给我一些关于如何解决我想做的事情的提示，因为我更有可能以这种方式学习。

Answer 1

您希望在遇到特定条件时break退出循环。

if condition:
            # do something
            break # brings you out of the loop

Answer 2

如果有人在 2020 年搜索这个：

import random

n = random.randint(1, 10)

guess_count = 0

guess_limit = 2      #actual try count will be 3

while guess_count <= guess_limit :

    guess = int(input("Enter an integer from 1 to 10: "))

    guess_count += 1
    
    if guess == n:

        print("you guessed it in ", guess_count,"Guesses")

        break
    
else:

    print ("Sorry, the correct number is", n)