Question

学习如何编写JPA查询。请告诉我是否可以更有效地编写以下查询，可以在单个选择语句中。可能是一个加入，但不知道该怎么做。

class Relationship {

  @ManyToOne
  public String relationshipType;  //can be MANAGER, CUSTOMER etc

  @ManyToOne
  public Party partyFrom; // a person who has a relation

  @ManyToOne
  public Party partyTo; // a group a person relate to
}

查询：

        String sql = "";
        sql = "select rel.partyTo";
        sql += " from Relationship rel";
        sql += " where rel.partyFrom = :partyFrom";
        sql += " and rel.relationshipType= :typeName";
        Query query = Organization.em().createQuery(sql);
        query.setParameter("partyFrom", mgr1);
        query.setParameter("typeName", "MANAGER");
        List<Party> orgList = query.getResultList();

        String sql2 = "";
        sql2 = "select rel.partyFrom";
        sql2 += " from Relationship rel";
        sql2 += " where rel.partyTo = :partyToList";
        sql2 += " and rel.relationshipType = :typeName2";
        Query query2 = Organization.em().createQuery(sql2);
        query2.setParameter("partyToList", orgList);
        query2.setParameter("typeName2", "CUSTOMER");
        List<Party> personList2 = query2.getResultList();

两个查询都有效。查询1返回一个组列表，其中person（mgr1）与MANAGER具有关系。查询2将它们所有的CUSTOMER人员返回给查询1返回的组。实际上，我得到一个他们所属的人员列表（客户），其中Person（mgr1）与MANAGER关系相同的组。

是否可以将它们组合成单个sql语句，因此可能只有一个db访问权限？

Answer 1

您实际上将一个查询嵌套在另一个查询中，并使用“where in”子句指定外部查询应从内部查询中获取客户。

select rel2.partyFrom
from Relationship rel2
where rel2.relationshipType = :typeName2 /* customer */
and rel2.partyTo.id in 
      (select rel.partyTo.id
      from Relationship rel
      where rel.partyFrom = :partyFrom
      and rel.relationshipType = :typeName)

您的通话代码会像以前一样通过typeName，typeName2和partyFrom参数。由于数据来自subselect（内部查询。）

，因此不需要PartyTo参数

你可以使用自连接实现相同的功能，使用where子句过滤左侧的管理器，使用右侧的客户，但使用'in'子句在语义上更清晰。

编辑：我将.id添加到subselect中，我认为这是必需的。

Answer 2

这不是问题的答案，但如果有人在使用JPQL调查Spring Data JPA中的@OneToMany关系时帮助其他人，因为问题与JPA有关，所以想分享我的2美分，提前道歉

@Entity
@Table(name = "MY_CAR")
public class MyCar {

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;

@Column(name = "DESCRIPTION")
private String description;

@Column(name = "MY_CAR_NUMBER")
private String myCarNumber;

@Column(name = "RELEASE_DATE")
private Date releaseDate;

@OneToMany(cascade = { CascadeType.ALL })
@JoinTable(name = "MY_CAR_VEHICLE_SERIES", joinColumns = @JoinColumn(name = "MY_CAR_ID "), inverseJoinColumns = @JoinColumn(name = "VEHICLE_SERIES_ID"))
private Set<VehicleSeries> vehicleSeries;
public MyCar() {
    super();
    vehicleSeries = new HashSet<VehicleSeries>();
}
// set and get method goes here


@Entity
@Table(name = "VEHICLE_SERIES ")
public class VehicleSeries {

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;

@Column(name = "SERIES_NUMBER")
private String seriesNumber;

@OneToMany(cascade = { CascadeType.ALL })
@JoinTable(name = "VEHICLE_SERIES_BODY_TYPE", joinColumns = @JoinColumn(name = "VEHICLE_SERIES_ID"), inverseJoinColumns = @JoinColumn(name = "BODY_TYPE_ID"))
private Set<BodyType> bodyTypes;
public VehicleSeries() {
    super();
    bodyTypes = new HashSet<BodyType>();
}
// set and get method goes here


@Entity
@Table(name = "BODY_TYPE ")
public class BodyType implements Serializable {

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;

@Column(name = "NAME")
private String name;
// set and get method goes here


public interface MyCarRepository extends JpaRepository<MyCar, Long> {
public Set<MyCar> findAllByOrderByIdAsc();

@Query(value = "select distinct myCar from MyCar myCar "
        + "join myCar.vehicleSeries as vs join vs.bodyTypes as bt where vs.seriesNumber like %:searchMyCar% "
        + "or lower(bt.name) like lower(:searchMyCar) or myCar.bulletinId like %:searchMyCar% "
        + "or lower(myCar.description) like lower(:searchMyCar) "
        + "or myCar.bulletinNumber like %:searchMyCar% order by myCar.id asc")
public Set<MyCar> searchByMyCar(@Param("searchMyCar") String searchMyCar);

}

表格中的一些数据，如

从Vehicle_Series

中选择*

ID      SERIES_NUMBER  
1       Yaris
2       Corolla

从Body_Type

中选择*

ID      NAME  
1       Compact
2       Convertible 
3       Sedan

如何编写JPA查询

2 个答案: