Hibernate/JPQL:如何基于对子项的查询将所有子项加载到父项中 [英] Hibernate/JPQL: How to load parent with all childs based on query on a child

问题描述

所以我们提供的服务简化了这两个实体

So we have a service with simplified these two entities

@Entity
public class Ticket {
/* simplified*/

  @OneToMany(fetch = FetchType.LAZY, mappedBy = "ticket", cascade = CascadeType.ALL, orphanRemoval = true)
  private Set<Grant> grants = new HashSet<>();
}

@Entity
public class Grant {
/* simplified*/

  @NotNull
  @ManyToOne(fetch = LAZY)
  @JoinColumn(name = UsageGrant.FK_TICKET, nullable = false)
  private Ticket ticket;

  @NotNull
  @Column(name = "specialNumber", nullable = false)
  private Integer specialNumber;
}

我想查询一个查询，该查询选择所有包含带有特定"specialNumber"赠款的票证.我要抓住的是，我想让票证与所有赠款一起退还，而不仅仅是一个匹配项.我尝试过

I'd like to have a query that selects all tickets that contain a grant with a specific "specialNumber". The catch is that I want to have the ticket returned with all grants, not only the one matching. I tried it with

@Repository
public interface TicketRepository extends JpaRepository<Ticket, String> {

@Query("SELECT DISTINCT ti FROM Ticket ti JOIN FETCH ti.grants g WHERE 
g.specialNumber = :specialNumber "
  )
  List<Ticket> findBySpecialNumberAndLoadAllGrantsOnTicket(
      @NotNull @Param("specialNumber") Integer specialNumber);
}

但是这给了我匹配的那个.我是否需要将其分为两个查询? Criteria API也无济于事，因为那里也不支持RIGHT JOIN.

but this gives me just the matching one. Do I need to split it up into two queries? Criteria API also doesn't help, because RIGHT JOIN is also not supported there.

更新

我可以做到

SELECT g FROM Grant g LEFT JOIN FETCH g.ticket ti JOIN FETCH ti.grants WHERE g.specialNumber = :specialNumber

，然后使用g.getTicket()访问票证.结果查询看起来很疯狂，我不确定这是否是一个聪明的方法.

and accessing the ticket with g.getTicket(). The resulting query looks crazy and I'm not sure if this is a clever approach at all.

推荐答案

您可以使用@EntityGraph来获取grants，并使用JPA方法查询以供specialNumber

You can use @EntityGraph for fetch grants and query using JPA method for select by specialNumber

@EntityGraph(attributePaths = {"grants"})
public List<Ticket> findByGrantsSpecialNumber(Integer specialNumber);

或

您可以使用@NamedEntityGraph

@NamedEntityGraph(name = "Ticket.Grants", attributeNodes = { @NamedAttributeNode("grants") })
public class Ticket {

和

@EntityGraph(value = "Ticket.Grants", type = EntityGraphType.LOAD)
public List<Ticket> findByGrantsSpecialNumber(Integer specialNumber);

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