BaseRepository
@NoRepositoryBean
public interface BaseRepository<T extends BaseEntity, ID extends Serializable>
extends JpaRepository<T, ID>, JpaSpecificationExecutor<T> {
T findByIdAndDeleteStatusFalse(ID id);
}
BaseServiceImpl
@Transactional(readOnly = true)
public abstract class BaseServiceImpl<T extends BaseEntity, ID extends Serializable> implements BaseService<T, ID> {
@Autowired
protected BaseRepository<T, ID> baseRepository;
}
applicationContext.xml中的jpa配置
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory"/>
<property name="jpaDialect">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaDialect" />
</property>
</bean>
<bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="dataSource" ref="dataSource" />
<property name="packagesToScan" value="com.coderbike.entity, com.coderbike.core.entity"/>
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
<property name="generateDdl" value="true" />
<property name="showSql" value="false"/>
<property name="databasePlatform" value="org.hibernate.dialect.MySQLDialect"/>
<property name="database" value="MYSQL"/>
</bean>
</property>
<property name="jpaProperties">
<props>
<prop key="hibernate.ejb.naming_strategy">org.hibernate.cfg.ImprovedNamingStrategy</prop>
<prop key="hibernate.show_sql">false</prop>
<prop key="hibernate.format_sql">false</prop>
</props>
</property>
</bean>
<!-- annotation transaction -->
<tx:annotation-driven transaction-manager="transactionManager" proxy-target-class="true" />
<!-- scan repository package -->
<jpa:repositories base-package="com.coderbike.dao.jpa, com.coderbike.core.repository"
repository-impl-postfix="Impl"
transaction-manager-ref="transactionManager"
entity-manager-factory-ref="entityManagerFactory" />
启动tomcat时,出现错误
Caused by: org.springframework.beans.factory.NoSuchBeanDefinitionException: No qualifying bean of type [com.coderbike.core.repository.BaseRepository] found for dependency [com.coderbike.core.repository.BaseRepository<com.coderbike.entity.User, java.lang.Integer>]: 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.support.DefaultListableBeanFactory.raiseNoMatchingBeanFound(DefaultListableBeanFactory.java:1406)
at org.springframework.beans.factory.support.DefaultListableBeanFactory.doResolveDependency(DefaultListableBeanFactory.java:1057)
at org.springframework.beans.factory.support.DefaultListableBeanFactory.resolveDependency(DefaultListableBeanFactory.java:1019)
at org.springframework.beans.factory.annotation.AutowiredAnnotationBeanPostProcessor$AutowiredFieldElement.inject(AutowiredAnnotationBeanPostProcessor.java:566)
... 60 more
我在网上搜索了很长时间。但最流行的场景是创建BaseRepositoryImpl并自定义扩展JpaRepositoryFactory的BaseRepositoryFactory。我可以使用自动装配而不是上面的方案。
答案 0 :(得分:1)
@NoRepositoryBean的概念不是为您创建特定存储库接口的代理实例。 here有很好的解释。
这意味着您可以扩展此BaseRepository接口并提供和实现您自己的接口。完成后,在xml中声明该bean。
答案 1 :(得分:0)
这意味着没有可用于实现该依赖关系的BaseRepository bean。
为BaseRepository使用注释@Component或@Repository,Spring将为您工作。