参数化类上的java Spring自动连接问题

1 年，8 月 Questions & Answers 1238

我已经为这个问题挣扎了一段时间，浏览了很多文章，但没有找到解决方案。感谢您对以下问题的帮助

我需要能够在服务类中自动连接EntityManager，但会引发异常。默认构造函数可能因为类型擦除而出现问题，因此我尝试使用带参数的构造函数来设置类型。如何自动连接用户类型的EntityManager

public interface IEntityManager<T extends IDomain<ID>, ID extends Serializable> { T findById(ID id, boolean lock); T save(T entity); void delete(T entity); } public class EntityManager<T extends IDomain<ID>, ID extends Serializable> implements IEntityManager<T, ID> { private Class<T> entity; @Autowired(required=true) private SessionFactory sessionFactory; /* @SuppressWarnings("unchecked") public EntityManager() { this.entity = (Class<T>) ((ParameterizedType) getClass() .getGenericSuperclass()).getActualTypeArguments()[0]; } */ @Autowired(required=true) public EntityManager(Class<T> entity) { this.entity = entity; } } @Service("UserService") @Transactional public class UserServiceImpl implements IUserService { @Autowired EntityManager<User, Integer> entityManager; }

以下是例外：

Exception in thread "main" org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'UserService': Injection of autowired dependencies failed; nested exception is org.springframework.beans.factory.BeanCreationException: Could not autowire field: com.test.dummy.persistence.manager.EntityManager com.test.dummy.service.UserServiceImpl.entityManager; nested exception is org.springframework.beans.factory.NoSuchBeanDefinitionException: No qualifying bean of type [com.test.dummy.persistence.manager.EntityManager] found for dependency: expected at least 1 bean which qualifies as autowire candidate for this dependency. Dependency annotations: {@org.springframework.beans.factory.annotation.Autowired(required=true)} at org.springframework.beans.factory.annotation.AutowiredAnnotationBeanPostProcessor.postProcessPropertyValues(AutowiredAnnotationBeanPostProcessor.java:292) at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.populateBean(AbstractAutowireCapableBeanFactory.java:1185) at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:537) at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:475)

# 2 楼答案

Spring无法推断它需要在EntityManager的构造函数中注入User.class，以满足UserServiceImpl中的深度

如果您使用的是Spring 4，则可以定义类型为EntityManager<User, Integer>的bean：

@Configuration
public class Config {
      @Bean 
      public EntityManager<User, Integer> userEntityManager() {
            new EntityManager(User.class);
      } 
}

编辑： 或者定义一个BeanDefinitionRegistryPostProcessor来检查需要EntityManager作为依赖项的bean，确定所需类型，为所需类型构造新的EntityManager bean，并将其添加到bean注册表

 @Component
 public class MyEntityManagerBeanDefinitionRegistryPostProcessor implements
                                                       BeanDefinitionRegistryPostProcessor {


   @Override
   public void postProcessBeanFactory(final ConfigurableListableBeanFactory beanFactory)
       throws BeansException {
     for (String beanName : beanFactory.getBeanDefinitionNames()) {
       final BeanDefinition beanDefinition = getOriginatingBeanDefinition(
           beanFactory.getBeanDefinition(beanName));
       final Class<?> beanClass = getClass(beanDefinition.getBeanClassName());
       if (beanClass != null) {
         ReflectionUtils.doWithFields(beanClass, new ReflectionUtils.FieldCallback() {
           public void doWith(Field field) throws IllegalArgumentException, IllegalAccessException {

             if (field.isAnnotationPresent(Autowired.class) || field.isAnnotationPresent(
                 Inject.class)) {
               if (field.getGenericType() instanceof ParameterizedType) {
                 final ParameterizedType parameterizedType = (ParameterizedType) field.getGenericType();
                 final Class<?> rawType = (Class) parameterizedType.getRawType();
                 if (rawType.equals(EntityManager.class)) {
                   final Class<?> typeArgument = (Class<?>) parameterizedType.getActualTypeArguments()[0];
                   beanFactory.registerSingleton(field.getName(), new EntityManager(typeArgument));
                 }
               }
             }
           }
         });
       }
     }
   }

   private Class<?> getClass(String className) throws BeansException {
     if (className != null) {
       try {
         return Class.forName(className);
       } catch (ClassNotFoundException e) {
         throw new BeanInitializationException("beanClass not found", e);
       }
     }
     return null;
   }

   private BeanDefinition getOriginatingBeanDefinition(BeanDefinition beanDefinition) {
     while(beanDefinition.getOriginatingBeanDefinition() != null) {
       beanDefinition = beanDefinition.getOriginatingBeanDefinition();
     }
     return beanDefinition;
   }


   @Override
   public void postProcessBeanDefinitionRegistry(BeanDefinitionRegistry registry) throws BeansException {  }
 }

最后，看起来您实际需要的是使用Spring Data Jpa，它本质上允许您通过定义如下接口来定义参数化DAO：

package com.myapp.repository; 

@Repository
public interface UserRepository extends JpaRepository<User, Long> {}

您可以启用它：

@Configuration
@EnableJpaRepositories(basePackages = "com.myapp.repository")
public class Config {


}

然后注入UserRepository：

@Autowired
private UserRepository userRepository;

Spring将注入一个基本的DAO实现

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参数化类上的java Spring自动连接问题

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# 1 楼答案

# 2 楼答案