Question

我需要加入四张桌子。这四个表是：

产品
服务
SERVICE_TYPE
service_duration

产品有服务，服务有类型和持续时间。

用户选择产品，类型，服务和服务的持续时间。

如果服务不适用于所选类型和持续时间，我希望能够获得具有较低优先级类型的服务（优先级是service_type表中的属性）。

在php（Laravel）中，我可以这样加入：

DB::table('product')
        ->join('product_service', 'product_service.product_id', '=', 'product.id')
        ->join('service', 'service.id', '=', 'product_service.service_id')
        ->join('service_type', 'service.type_id', '=', 'service_type.id')
        ->join('service_duration', 'service.duration_id', '=', 'service_duration.id')
        ->where('product.id', id)
        ->where('service_type.priority', '<', priority)
        ->where('service_duration.duration', duration)
        ->where('maintenance.active', 1)
        ->orderBy('service_type.priority', 'desc')
        ->select('service.*')
        ->first();

如何使用Hibernate Entity Criteria执行此操作？我想从产品方面加入，但最后选择服务。

我的关系定义如下：

产品类

 @Entity
 @Table(name = "product")
 public class product implements Serializable {
        @OneToMany(fetch = FetchType.LAZY, mappedBy = "id.product", cascade=CascadeType.ALL)
        private Set<ProductService> services = new HashSet<ProductService>();
  }

ProductService类：

@Entity
@Table(name = "product_service")
@AssociationOverrides({ @AssociationOverride(name = "id.service", joinColumns = @JoinColumn(name = "service_id")),
        @AssociationOverride(name = "id.product", joinColumns = @JoinColumn(name = "product_id"))})

public class ProductService  implements Serializable {

    private static final long serialVersionUID = 1L;

    @EmbeddedId
    private ProductServicePK id = new ProductServicePK();


    @Column(name = "price")
    private Float price;

    @Column(name = "code")
    private String code;
}

ProductServicePK类：

@Embeddable
public class ProductServicePK implements Serializable {

    private static final long serialVersionUID = 1L;

    @ManyToOne
    private Product product;

    @ManyToOne
    private Service service;
}

服务类：

   @Entity
@Table(name = "service")
public class Service implements Serializable {

    @Id
    @Column(name = "id")
    private Long id;

    @OneToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "type_id")
    private ServiceType type;

    @OneToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "duration_id")
    private ServiceDuration duration;
}

因此，“保持”引用另一个的对象是产品对象。因此，我不确定如何获得优先级低于所选产品的产品服务。

我想用条件或HQL来做这件事。

Answer 1

你的Hibernate Detached Criteria会喜欢：

DetachedCriteria criteria = DetachedCriteria.forClass(Product.class,     "product")
    .criteria.createAlias("product.product_service", "productService")
    .createAlias("productService.service","service")
    .createAlias("service.service_type","serviceType")
    .createAlias("service_duration","serviceDuration")
    .add(Restrictions.eq("product.id", id))
    .add(Restrictions.gt("serviceType.priority", priority))
    .add(Restrictions.eq("serviceDuration.duration", duration))
    .setProjection(add your productService projection here)
    getExecutableCriteria(session).SetMaxResults(1) ;

但我不明白你的意思，为什么你在from query使用product代替product_service？，因为你需要product_service个详细信息。

您可以使用querycriteria.html，这是doc的文档 - ＆gt; https://docs.jboss.org/hibernate/orm/3.3/reference/en/html/querycriteria.html 或者您可以使用hql - ＆gt; https://docs.jboss.org/hibernate/orm/3.3/reference/en/html/queryhql.html来获得结果。

Answer 2

显然，JPA Criteria API不支持使用复合PK模式时所需的内容。我发现这提到here。所以我发现最好的解决方案是使用HQL：

Query q = session.createQuery("select s "
            + "from Service s "
            + "join s.productService as ps "
            + "join s.type as t "
            + "where s.duration = :durationId "
            + "and ps.id.product.id = :productId "
            + "and t.priority <= :priorityValue "
            + "order by t.priority desc");
    q.setLong("durationId", durationId);
    q.setInteger("priorityValue", priorityValue);
    q.setLong("productId", productId);

    return (Service) q.setMaxResults(1).uniqueResult();

如何使用休眠

2 个答案: