Java 如何使用hibernate连接四个表
我有四张桌子需要加入。这四个表格是:Java 如何使用hibernate连接四个表,java,php,hibernate,laravel,Java,Php,Hibernate,Laravel,我有四张桌子需要加入。这四个表格是: 产品 服务 服务类型 服务时间 产品有服务,服务有类型和持续时间 用户选择产品、类型、服务和服务期限 如果服务在所选类型和持续时间内不可用,我希望能够获得优先级较低的服务(优先级是服务类型表中的属性) 在php(Laravel)中,我可以这样加入: DB::table('product') ->join('product_service', 'product_service.product_id', '=', 'product.id')
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();
ProductServicePK类:
@Embeddable
public class ProductServicePK implements Serializable {
private static final long serialVersionUID = 1L;
@ManyToOne
private Product product;
@ManyToOne
private Service service;
}
服务类别:
@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,“持有”对另一个对象的引用的对象就是产品对象。因此,我不知道如何为优先级低于所选产品的产品获得服务
我想用criteria或HQL执行此操作。您的Hibernate分离标准如下:
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) ;
但是我不明白你的意思,为什么你在产品
上使用而不是产品服务
?,因为你需要产品服务
的详细信息
您可以使用querycriteria.html,这是它的文档->
或者您可以使用hql->来实现您的结果 显然,JPA标准API不支持我在使用复合PK模式时所需的内容。我在某处发现了这个。因此,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();
Query q=session.createQuery(“选择s”
+“来自服务s”
+“以ps身份加入s.productService”
+“将s.type作为t连接”
+“其中s.duration=:durationId”
+“和ps.id.product.id=:productId”
+ "和t.priority是的,您可以这样做。您的实体类是否已经使用@OneToOne
/@OneToMany
/@ManyToOne
关系进行了设置?Hibernate文档清楚地解释了如何处理关系。您看过Hibernate文档了吗?@deanclk是的,它们是。我不介意在其他时间发生这种情况我只是想得到相同的结果。@user3552551那么你可以使用我上面提到的DetachedCriteria。我想我是从产品方面开始的,因为这是定义多对多关系的地方。我将尝试你的解决方案,谢谢。
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) ;
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();