Java 持久化对象时没有可用的事务性EntityManager
当我试图持久化一个对象时,没有可用的事务EntityManager异常 我有一个具有以下配置类的Spring项目:Java 持久化对象时没有可用的事务性EntityManager,java,spring,jpa,persistent,Java,Spring,Jpa,Persistent,当我试图持久化一个对象时,没有可用的事务EntityManager异常 我有一个具有以下配置类的Spring项目: @Configuration @ComponentScan @EnableTransactionManagement @EnableJpaRepositories public class SpringConfiguration { @Bean(name = "entityManagerFactory") public LocalContainerEntityManagerFact
@Configuration
@ComponentScan
@EnableTransactionManagement
@EnableJpaRepositories
public class SpringConfiguration {
@Bean(name = "entityManagerFactory")
public LocalContainerEntityManagerFactoryBean emf(DataSource dataSource) {
LocalContainerEntityManagerFactoryBean emf =new LocalContainerEntityManagerFactoryBean();
emf.setDataSource(dataSource);
Properties jpaProperties = new Properties();
jpaProperties.put("hibernate.hbm2ddl.auto", "create-drop");
jpaProperties.put("hibernate.show_sql", "true");
emf.setJpaProperties(jpaProperties);
emf.setPackagesToScan(
new String[] {"ds.core.entity"});
emf.setJpaVendorAdapter(
new HibernateJpaVendorAdapter());
return emf;
}
@Bean(name = "transactionManager")
public PlatformTransactionManager transactionManager(EntityManagerFactory emf,DataSource dataSource) {
JpaTransactionManager tm =
new JpaTransactionManager();
tm.setEntityManagerFactory(emf);
tm.setDataSource(dataSource);
return tm;
}
和实体类:
@Entity
public class Type {
@Id @GeneratedValue
private Long id;
private String name;
}
和Jpa实现类:
@Repository
public class JpaTypeRepo implements TypeRepo {
@PersistenceContext
private EntityManager em;
@Override
public Type insertRecord(Type data) {
em.persist(data);
return data;
}
}
最后是一个带有Init()
的服务类:
只需将@Transactional注释添加到类:JpaTypeRepo 详情如下:
@Transactional
@Repository
public class JpaTypeRepo implements TypeRepo {
@PersistenceContext
private EntityManager em;
@Override
public Type insertRecord(Type data) {
em.persist(data);
return data;
}
}
@Transactional
@Repository
public class JpaTypeRepo implements TypeRepo {
@PersistenceContext
private EntityManager em;
@Override
public Type insertRecord(Type data) {
em.persist(data);
return data;
}
}