While循环2整数值

Question

我正在尝试编写一个函数，从用户那里获取2个int值，直到它们的总和是21，很简单！

其目的是不断提示用户传递2个int值，直到满足条件为止。我不确定代码在哪里中断，因为它在满足任何一个条件时都会停止，无论是true还是false。

def check_for_21(n1,n2):
    result = 0
    while True:
        while result != 21:
            try:
                n1 = int(input("Enter first number >> "))
                n2 = int(input("Enter second number >> "))

                if n1+n2 != 21:
                    print("You did not get to 21! ")
                else:
                    print("You got it! ")

            except:
                if n1+n2 == 21:
                    print("You got it! ")
            else:
                break

        break

Answer 1

这就是你要找的，你有多个逻辑错误！然而，这个想法是存在的，但格式错误。在这个程序中，它会一直运行，直到您输入两个数字，它们的总和是21

def check_for_21():
    while True:
        n1 = int(input("Enter first number >> "))
        n2 = int(input("Enter second number >> "))

        if n1+n2 != 21:
            print("You did not get to 21! ")
        else:
            print("You got it! ")
            break

check_for_21()

Answer 2

尝尝这个。这将满足您的要求。

a = 5
while (a<6):
        n1 = int(input("Enter first number >> "))
        n2 = int(input("Enter second number >> "))

        if n1+n2 == 21:
                print("You got it ")
                break
        else:
                print("You did not get to 21!  ")

Answer 3

接受输入，检查结果，如果得到21，则中断循环。

def check_for_21(n1,n2):
    result = 0
    while True:
        try:
            n1 = int(input("Enter first number >> "))
            n2 = int(input("Enter second number >> "))

            if n1+n2 != 21:
                print("You did not get to 21! ")
            else:
                print("You got it! ")
                break

        except:
            pass