Spring Java Configuration: Create Bean From Dynamic Class Name

Question

My situation is that the class name of a bean can be configured by an external properties-file. It's fine to do this with XML

  <bean id="myService" class="${app.serviceClass}" />

and properties-file

app.serviceClass=com.example.GreatestClassThereIs

I tried to convert it to Java using a BeanFactoryPostProcessor:

@Configuration
public class MyConfiguration implements BeanFactoryPostProcessor {

  @Value("${app.serviceClass}")
  private String serviceClassName;

  @Override
  public void postProcessBeanFactory(ConfigurableListableBeanFactory beanFactory) throws BeansException {
    logger.warn("let's register bean of class " + serviceClassName + "...");
    GenericBeanDefinition beanDefinition = new GenericBeanDefinition();
    beanDefinition.setBeanClassName(serviceClassName);

    BeanDefinitionRegistry registry = (BeanDefinitionRegistry) beanFactory;
    registry.registerBeanDefinition("myService", beanDefinition);
    logger.warn("done " + beanDefinition);
  }
}

The problem is that the lifecycle of Spring hasn't yet set handled the @Value and serviceClassName is null. How do I get the property in there?

Not sure what you are trying to achieve. You have a @Configuration class which is actually a BeanFactoryPostProcessor. Seems strange to me. Defining a @Bean and use the Spring ClassUtils to create an instance of that bean. Else registering a RootBeanDefinition with ${app.serviceClass} might also work. — M. Deinum
– M. Deinum, Commented May 13, 2014 at 14:33
@M.Deinum Well, the source in my question is just the important stuff for this question. I don't want to write off-topic stuff that may cause more confusion. I will check your two suggestions. Would both cause Spring to do DI and such on the bean? — sjngm
– sjngm, Commented May 13, 2014 at 16:11
@SotiriosDelimanolis I completely fail to understand what the supposedly duplicate question is about. I can see that an answer suggests to use BeanDefinitionRegistryPostProcessor, which I haven't heard of yet, but the rest is so confusing I can't even argue whether or not my question is a duplicate ;) — sjngm
– sjngm, Commented May 14, 2014 at 6:30

sjngm · Accepted Answer · 2014-05-14 07:08:59Z

4

Why not simply define a new Bean in your ApplicationContext by using Class.forName() from the @Value injected into the @Configuration?

So something like this:

@Configuration
public class MyConfiguration
{

     @Value("${app.serviceClass}")
     private String serviceClassName;

     @Bean
     public Object myService()
     {
          return Class.forName(serviceClassName).newInstance();
     }
}

EDIT by sjngm (for better readability than in the comment):

     @Bean
     public MyInterface myService()
     {
        Class<?> serviceClass = Class.forName(serviceClassName);
        MyInterface service = MyInterface.class.cast(serviceClass.newInstance());
        return service;
     }

edited May 14, 2014 at 7:08

sjngm

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answered May 13, 2014 at 14:32

Rob Lockwood-Blake

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2 Comments

sjngm Over a year ago

I guess that when implementing against interfaces the return type of myService can be the interface. This is important for DI of this bean in other places.

Sushant Gupta Over a year ago

How do we implement this if we do not have serviceClassName during application startup?

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