Spring Boot EntityManager不持久化关系实体

Question

我尝试保留@OneToMany，但出现错误：

  

插入

into
    booking
    (booking_date, cancel_date) 
values
    (?, ?)

     

将参数[1]绑定为[DATE]-[2018-07-20]

     

将参数[2]绑定为[DATE]-[null]

     

org.hibernate.SQL：

insert 
into
    category_booking
    (name) 
values
    (?)

     

com.arjuna.ats.jta：ARJUNA016060：TransactionImple.enlistResource-   捕获：XAException.XAER_INVAL

     

o.h.engine.jdbc.spi.SqlExceptionHelper：SQL错误：0，SQLState：   空

     

o.h.engine.jdbc.spi.SqlExceptionHelper：   ConnectionImple.registerDatabase-ARJUNA017017：资源清单   失败

     

o.h.engine.jdbc.spi.SqlExceptionHelper：SQL错误：0，SQLState：   空

     

o.h.engine.jdbc.spi.SqlExceptionHelper：ARJUNA017005：错误   发生在关闭期间：

     

com.arjuna.ats.jta：ARJUNA016029：   SynchronizationImple.afterCompletion-失败

     

org.hibernate.resource.transaction.backend.jta.internal.synchronization.RegisteredSynchronization@5c66658b   例外

     

org.hibernate.exception.GenericJDBCException：无法发布JDBC   连接

这是我的实体：

@Entity
@Table(name="booking")
public class Booking implements Serializable {  
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    @Column(name="id")
    private Long id;
    @Column(name="booking_date")
    private Date bookingDate;   
    @Column(name="cancel_date")
    private Date cancelDate;    
    @OneToMany(cascade = CascadeType.ALL, orphanRemoval = true)
    @JoinColumn(name = "booking_id")
    private List<CategoriesBooking> bookingInfos = new ArrayList<>();   
}
@Entity
@Table(name="category_booking")
public class CategoriesBooking implements Serializable{ 
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    @Column(name="id")
    private Long id;
    @Column(name = "name")  
    private String name;
}

持久功能

@Repository
public class ServicesRepos{
    @PersistenceContext
    protected EntityManager entityManager;

    public AbstractServices() {     
    }

    public EntityManager getEm() {
        return entityManager;
    }   

    @Transactional
    @SuppressWarnings("unchecked")
    public Object add(Object entity){
        try {
            EntityManager em = getEm();
            em.persist(entity);
            return entity;
        } catch (Exception e) {
            entityManager.getTransaction().rollback();          
        }
        return null;
    }
}

和我的输入数据：

{
 bookingdate: 2018-07-20,
 bookingInfos: [{
   name: "test for persist"
}]
}