python - 如何从类中的另一个函数(方法)调用一个函数(方法)？

Question

我想创建 switch case，但不知道如何从一个函数调用另一个函数。例如，在这段代码中，我想在按下“O”时调用 OrderModule()。

实际上我已经用Java完成了整个程序，也许有人知道，如何更轻松地转换它而不需要重新编写它？

class CMS:

    def MainMenu():
        print("|----Welcome to Catering Management System----|")

        print("|[O]Order")
        print("|[R]Report")
        print("|[P]Payment")
        print("|[E]Exit")
        print("|")
        print("|----Select module---- \n")
        moduleSelect = raw_input()
        if moduleSelect == "o" or moduleSelect == "O":
            ---
        if moduleSelect == "r" or moduleSelect == "R":
            ---
        if moduleSelect == "P" or moduleSelect == "P":
            ---
        if moduleSelect == "e" or moduleSelect == "E":
            ---
        else:
            print("Error")
    MainMenu()
    def OrderModule():
        print("|[F]Place orders for food")
        print("|[S]Place orders for other services")
        print("|[M]Return to Main Menu")
    OrderModule()

Answer 1

这是交易。为了让您理解 Python，我将稍微重构一下代码，也许还会提供一些有关设计模式的好技巧。

请考虑这个示例过于简单且有些过度，它的唯一目的是提高您的新兴开发技能。

首先，熟悉一下 Strategy Design Pattern 是件好事。 ，这对于此类任务非常有用(我个人认为)。之后，您可以创建一个基本模块类及其策略。 注意self(表示对象本身实例的变量)如何作为第一个参数显式传递给类方法:

class SystemModule():
    strategy = None

    def __init__(self, strategy=None):
        '''
        Strategies of this base class should not be created
        as stand-alone instances (don't do it in real-world!).
        Instantiate base class with strategy of your choosing
        '''
        if type(self) is not SystemModule:
            raise Exception("Strategy cannot be instantiated by itself!")
        if strategy:
            self.strategy = strategy()

    def show_menu(self):
        '''
        Except, if method is called without applied strategy
        '''
        if self.strategy:
            self.strategy.show_menu()
        else:
            raise NotImplementedError('No strategy applied!')


class OrderModule(SystemModule):
    def show_menu(self):
        '''
        Strings joined by new line are more elegant
        than multiple `print` statements
        '''
        print('\n'.join([
            "|[F]Place orders for food",
            "|[S]Place orders for other services",
            "|[M]Return to Main Menu",
        ]))


class ReportModule(SystemModule):
    def show_menu(self):
        print('---')


class PaymentModule(SystemModule):
    def show_menu(self):
        print('---')

此处，OrderModule、ReportModule 和 PaymentModule 可以定义为一等函数，但在本示例中，类更为明显. 接下来，创建应用程序的主类:

class CMS():
    '''
    Options presented as dictionary items to avoid ugly
    multiple `if-elif` construction
    '''
    MAIN_MENU_OPTIONS = {
        'o': OrderModule, 'r': ReportModule, 'p': PaymentModule,
    }

    def main_menu(self):
        print('\n'.join([
            "|----Welcome to Catering Management System----|", "|",
            "|[O]Order", "|[R]Report", "|[P]Payment", "|[E]Exit",
            "|", "|----Select module----",
        ]))

        # `raw_input` renamed to `input` in Python 3,
        # so use `raw_input()` for second version. Also,
        # `lower()` is used to eliminate case-sensitive
        # checks you had.
        module_select = input().lower()

        # If user selected exit, be sure to close app
        # straight away, without further unnecessary processing 
        if module_select == 'e':
            print('Goodbye!')
            import sys
            sys.exit(0)

        # Perform dictionary lookup for entered key, and set that
        # key's value as desired strategy for `SystemModule` class
        if module_select in self.MAIN_MENU_OPTIONS:
            strategy = SystemModule(
                strategy=self.MAIN_MENU_OPTIONS[module_select])

            # Base class calls appropriate method of strategy class
            return strategy.show_menu()
        else:
            print('Please, select a correct module')

为了使整个事情顺利进行，文件末尾有一个简单的启动器:

if __name__ == "__main__":
    cms = CMS()
    cms.main_menu()

给你。我真的希望这段代码能够帮助您更深入地了解 Python :)干杯!