在动态创建的类中实例化Spring Bean [英] Instantiating spring beans in dynamically created classes

问题描述

我正在动态创建包含Spring bean的类，但是并没有实例化或初始化bean，而是将它们保留为空.

I am dynamically creating classes which contain spring beans, however the beans are not getting instantiated or initialised, leaving them as null.

如何确保动态创建的类正确创建了所有其spring bean?

How do I make sure that a dynamically created class creates all of its spring beans properly?

这是我动态创建类的方式:

This is how I am dynamically creating the class:

Class ctransform;
try {
    ctransform = Class.forName(strClassName);
    Method handleRequestMethod = findHandleRequestMethod(ctransform);
    if (handleRequestMethod != null) {
        return (Message<?>) handleRequestMethod.invoke(ctransform.newInstance(), message);
            }
    }

这会将ctransform(类型为strClassName)内的所有spring bean对象保留为空.

This leaves all spring bean objects within ctransform (of type strClassName) as null.

推荐答案

每当实例化类时，它们都是不是春季管理的. Spring必须实例化类，以便它可以注入其依赖项.除了使用@Configurable和<context:load-time-weaver/>的情况例外，但这更像是一种hack，我建议您反对.

Whenever you instantiate classes, they are not spring-managed. Spring has to instantiate classes so that it can inject their dependencies. This with the exception of the case when you use @Configurable and <context:load-time-weaver/>, but this is more of a hack and I'd suggest against it.

相反:

• 使范围为prototype
的bean
• 获取ApplicationContext(在网络应用中，这是通过WebApplicationContextUtils.getRequiredWebApplicationContext(servletContext)完成的)
• 如果未注册类(并且我假设它们未注册)，请尝试强制转换为StaticApplicationContext(我不确定这是否可行)，然后调用registerPrototype(..)在上下文中注册您的类.如果这不起作用，请使用GenericContext及其registerBeanDefinition(..)
• 使用appContext.getBeansOfType(yourclass)获取与您的类型匹配的所有实例；或者，如果您刚刚注册它并知道它的名称-请仅使用appContext.getBean(name)
• 确定哪一个适用.通常，Map中只有一个条目，请使用它.

• make the bean of scope prototype
• obtain the ApplicationContext (in a web-app this is done via WebApplicationContextUtils.getRequiredWebApplicationContext(servletContext))
• if the classes are not registered (and I assume they are not), try casting to StaticApplicationContext (I'm not sure this will work), and call registerPrototype(..) to register your classes in the context. If this doesn't work, use GenericContext and its registerBeanDefinition(..)
• get all the instances that match your type, using appContext.getBeansOfType(yourclass); or if you just registered it and know its name - use just appContext.getBean(name)
• decide which one is applicable. Usually you will have only one entry in the Map, so use it.

但是我通常会避免对弹跳豆进行反思-应该有另一种方法来实现这一目标.

But I would generally avoid reflection on spring beans - there should be another way to achieve the goal.

更新:我只是想到了一个更简单的解决方案，该解决方案在不需要注册Bean的情况下也可以使用-即，您动态生成的类不会注入任何其他动态生成的类中课:

Update: I just thought of an easier solution, that will work if you don't need to register the beans - i.e. that your dynamically generated classes won't be injected in any other dynamically generated class:

// using WebApplicationContextUtils, for example
ApplicationContext appContext = getApplicationContext(); 
Object dynamicBeanInstance = createDyamicBeanInstance(); // your method here
appContext.getAutowireCapableBeanFactory().autowireBean(dynamicBeanInsatnce);

您将设置依赖项，而无需将新类注册为bean.

And you will have your dependencies set, without having your new class registered as a bean.

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