我有两张表如下: // TmCategory表
public class TmCategory implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private Long id;
@Column(name="CATEGORY_NAME")
private String categoryName;
@Column(name="OWNER_ID")
private BigInteger ownerId;
//bi-directional many-to-one association to TmCategoryRate
@OneToMany(mappedBy="tmCategory", cascade = CascadeType.ALL, fetch = FetchType.EAGER)
@javax.persistence.OrderBy("startDate ASC")
private Set<TmCategoryRate> tmCategoryRates;
//getters and setters
}
// TmCategoryRates表
@Entity
@Table(name="tm_category_rates")
public class TmCategoryRate implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private Long id;
@Column(name="CREATED_BY")
private BigInteger createdBy;
@Column(name="CREATED_DATE")
private Timestamp createdDate;
@Column(name="END_DATE")
private Timestamp endDate;
@Column(name="RATE")
private Double rate;
@Column(name="RATE_TYPE")
private String rateType;
@Column(name="START_DATE")
private Timestamp startDate;
//bi-directional many-to-one association to TmCategory
@ManyToOne
@JoinColumn(name="CATEGORY_ID")
private TmCategory tmCategory;
//getters and setters.....
}
以下是用于获取类别的分离标准
DetachedCriteria criteria = DetachedCriteria.forClass(TmCategory.class);
criteria.add(Restrictions.eq("id", catId));
DetachedCriteria criteria1 =criteria.createCriteria("tmCategoryRates");
criteria1.addOrder(Order.asc("id"));
List<TmCategory> categories = getHibernateTemplate().findByCriteria(criteria);
这里我试图按照ID的升序对tmCategoryRates进行排序。 如果此类别至少有一个tmCategoryRates,我可以获得类别。如果没有可用的tmCategoryRates,则返回null而不是使用tmCategoryRates返回类别。
答案 0 :(得分:0)
如果该类别可能没有费率,则需要左连接。目前尚不清楚为什么你在这里使用DetachedCriteria,但是从Hibernate docs看这个例子
List cats = session.createCriteria( Cat.class )
.createAlias("mate", "mt",
Criteria.LEFT_JOIN,Restrictions.like("mt.name", "good%") )
.addOrder(Order.asc("mt.age"))
.list();