是否有可能阻止在Python另一个函数的循环？

Question

我想，随着主菜单启动一个程序，关闭功能不同的分支。在每个函数结束时，我希望用户要问'你想回到主菜单？如果用户说没有我想要的程序通过停止主循环终止（我不希望使用sys.exit（）或任何类似）。示例代码：

#Issue is almost at the bottom
#Feel free to comment on the rest of the code as well,
#Always looking to improve
def main():
    loop = True
    while loop:
        print('''MENU CHOICE''')
        print('''1: go here''')
        print('''2: go there''')
        print('''3: You get the point''')
        print('''0: Terminate program''')
        print()

        try:
            answer = int(input('''I want to go to program: '''))
        except:
            print('''Not a valid menu choice, please try again''')
            print()

        if answer != 1 and answer != 2 and answer != 3 and answer != 0:
            print('''Not a valid menu choice, please try again''')
            print()
        elif answer == 1:
            program1()
        elif answer == 2:
            program2()
        elif answer == 3:
            program3()
        else:
            loop = False

def program1():
    print('''This is program 1''')
    itdontwork = input('''Do you want to go back to the menu? Y/N''')

    if itdontwork == 'Y' or itdontwork == 'y':
        print()
    else:
        print('''SHUTTING DOWN''')
        loop = False #Here is the issue

#The rest of the programs would be the same

main()

Answer 1

最简单的方法就是让你的变量循环全球

loop = True
def main():
    global loop
    while loop:
        print('''MENU CHOICE''')
        print('''1: go here''')
        print('''2: go there''')
        print('''3: You get the point''')
        print('''0: Terminate program''')
        print()

        try:
            answer = int(input('''I want to go to program: '''))
        except:
            print('''Not a valid menu choice, please try again''')
            print()

        if answer != 1 and answer != 2 and answer != 3 and answer != 0:
            print('''Not a valid menu choice, please try again''')
            print()
        elif answer == 1:
            program1()
        elif answer == 2:
            program2()
        elif answer == 3:
            program3()
        else:
            loop = False

def program1():
    global loop
    print('''This is program 1''')
    itdontwork = input('''Do you want to go back to the menu? Y/N''')

    if itdontwork == 'Y' or itdontwork == 'y':
        print()
    else:
        print('''SHUTTING DOWN''')
        loop = False #Here is the issue

#The rest of the programs would be the same

main()

全球将允许你有相同的W / R变数无处不在。如果没有全球的所有变量本地。

Answer 2

问题是，你正在试图更改在program1功能范围之外定义的变量。 loop被内部main定义因此仅main可以访问它。有几种方法来解决这个问题，你可以声明loop外（使其成为一个全球性），或只是让你的program1返回一个boolean值以调用函数，例如：

def main():
    loop = True
    while loop:
        loop = program1()

def program1():
    itdontwork = input('''Do you want to go back to the menu? Y/N''')
    if itdontwork == 'Y' or itdontwork == 'y':
        print()
    else:
        print('''SHUTTING DOWN''')
        return False

Answer 3

我想你想尝试做的是有不同的程序返回一个值。

问题是，该功能得到执行，但返回什么。你的主要功能没有返回值;因此，循环变量不能进行更新，以打破循环。

#Issue is almost at the bottom
#Feel free to comment on the rest of the code as well,
#Always looking to improve
def main():
    loop = True
    while loop:
        print('''MENU CHOICE''')
        print('''1: go here''')
        print('''2: go there''')
        print('''3: You get the point''')
        print('''0: Terminate program''')
        print()

        try:
            answer = int(input('''I want to go to program: '''))
        except:
            print('''Not a valid menu choice, please try again''')
            print()

        if answer != 1 and answer != 2 and answer != 3 and answer != 0:
            print('''Not a valid menu choice, please try again''')
            print()
        elif answer == 1:
            return program1() # Return the output of this function
        elif answer == 2:
            return program2() # Return the output of this function
        elif answer == 3:
            return program3() # Return the output of this function
        else:
            loop = False

def program1():
    print('''This is program 1''')
    itdontwork = input('''Do you want to go back to the menu? Y/N''')

    if itdontwork == 'Y' or itdontwork == 'y':
        print()
    else:
        print('''SHUTTING DOWN''')
        return False # Return the output of this function

#The rest of the programs would be the same

main()

Answer 4

...
        elif answer == 1:
            loop = program1()
        elif answer == 2:
            loop = program2()
        elif answer == 3:
            loop = program3()
        else:
            loop = False

def program1():
    print('''This is program 1''')
    itdontwork = input('''Do you want to go back to the menu? Y/N''')

    if itdontwork == 'Y' or itdontwork == 'y':
        print()
        return 1 # back to menu...
    else:
        print('''SHUTTING DOWN''')
        return 0

这将让你想要的，你从被调用函数需要在循环中的数据。

Answer 5

你可以在它raise ValueError('Words and stuff')然后陷阱。

if answer != 1 and answer != 2 and answer != 3 and answer != 0:
    print('''Not a valid menu choice, please try again''')
    print()
elif answer == 1:
    try:
        program1()
    except ValueError:
        break
elif answer == 2:
    try:
        program2()
    except ValueError:
        break
elif answer == 3:
    try:
        program3()
    except ValueError:
        break
else:
    loop = False


def program1():
    print('''This is program 1''')
    itdontwork = input('''Do you want to go back to the menu? Y/N''')

    if itdontwork == 'Y' or itdontwork == 'y':
        print()
    else:
        print('''SHUTTING DOWN''')
        # loop = False #Here is the issue
        raise ValueError('BOOM SHAKA LAKA!')