英文:
Load a class dynamically
问题
以下是翻译好的部分，不包括代码：
我有一个在启动时动态添加模块的软件。
启动发生在我的Python程序启动之后，所以我无法使用普通的“import DynModule”导入模块，而必须在运行时进行此操作。
因此，我想要做类似于以下的事情：
def load_module():try:import sitesite.addsitedir(r"C:\Users\...")from DynModule import Fooexcept (ModuleNotFoundError, ImportError) as e:logger.error("无法导入动态模块")class AccessClass(Process):def __init__(self):Process.__init__(self)logger.info("占位符类实例")def do_something(self):logger.info("占位符执行某些操作")else:class AccessClass(Process):def __init__(self):Process.__init__(self)logger.info("正确的类实例，用于动态加载的模块")def do_something(self):logger.info("实际执行某些操作")
然后在主程序中执行以下操作：
def action_on_load(self):from loader import load_module, AccessClassload_module()self._access = AccessClass()self._access.do_something() # 输出 "实际执行某些操作"
然而，由于AccessClass是在try-except子句内部定义的，因此我无法导入它。我该如何继续并使其正常工作？
英文:
I have a software that dynamically adds modules at startup.
The start happens after the start of my Python program, so I cannot import the modules with a normal "import DynModule", but have to do this during runtime.
Therefore I would like to do something like this:
def load_module():try:import sitesite.addsitedir(r"C:\Users\...")from DynModule import Fooexcept (ModuleNotFoundError, ImportError) as e:logger.error("Could not import the dynamic module")class AccessClass(Process):def __init__(self):Process.__init__(self)logger.info("Placeholder class instance")def do_something(self):logger.info("Placeholder does something")else:class AccessClass(Process):def __init__(self):Process.__init__(self)logger.info("The right class instance to use the dynamically loaded module")def do_something(self):logger.info("Actually does something")
And then do something like this in the main program:
def action_on_load(self):from loader import load_module, AccessClassload_module()self._access = AccessClass()self._access.do_something() # prints "Actually does something"
However, I cannot import AccessClass as it is defined inside the try-except clause.
How can I proceed and get this to work?
答案1
得分: 1
以下是您要翻译的内容：
问题不在于它在try语句中定义，而在于它在函数中定义，其中名称AccessClass是一个局部变量。您可以将名称声明为全局变量：
def load_module():global AccessClass...
因为class语句就像是一个复杂的赋值语句：它定义了一个类，然后将其绑定到一个名称上。不过，请注意，该类直到您实际调用函数时才会分配给该名称。在您的主程序中，您需要编写类似于以下内容：
def action_on_load(self):import loader # 还没有AccessClassloader.load_module() # 现在loader.AccessClass已定义。self._access = loader.AccessClass()self._access.do_something() # 输出“实际执行某些操作”
或者，您可以简单地返回该类：
def load_module():...return AccessClassdef action_on_load(self):from loader import load_moduleAccessClass = load_module()self._access = AccessClass()self._access.do_something() # 输出“实际执行某些操作”
注意：代码部分不需要翻译。
英文:
The problem isn't that its defined in the try statement, but that it's defined in the function, where the name AccessClass is a local variable. You can declare the name as global:
def load_module():global AccessClass...
because a class statement is like a fancy assignment statement: it defines a class, then binds it to a name. Note, though, that the class won't be assigned to the name until you actually call the function. In your main program, you'd need to write something like
def action_on_load(self):import loader # No AccessClass yetloader.load_module() # Now loader.AccessClass is defined.self._access = loader.AccessClass()self._access.do_something() # prints "Actually does something"
Or, you can simply return the class:
def load_module():...return AccessClassdef action_on_load(self):from loader import load_moduleAccessClass = load_module()self._access = AccessClass()self._access.do_something() # prints "Actually does something"
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