Question

我有以下两个具有一对多关系的pojo，我的查询是，现在可以通过SqlResultSetMapping来获取Com的第一类的结果，但是在Com内也有SubCom的引用，但由于我当前获得的SqlResultSetMapping输出是完美的，但它包含SubCom的null，因此请告知我如何修改SqlResultSetMapping以获取SubCom的值，我应该为SubCom设计单独的SqlResultSetMapping。

第一个Pojo如下所示

@SqlResultSetMapping(
        name = "firstQuery",
        classes = @ConstructorResult(
                targetClass = com.entity.Com.class,
                columns = {
                        @ColumnResult(name = "com_Id" , type = Integer.class),
                        @ColumnResult(name = "Cescription" , type = String.class),
                        @ColumnResult(name = "name" , type = String.class)}))




@Entity
@Data
@Table(name = "COM")
public class Com implements Serializable {

        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)

        @Column(name = "com_Id")
        private Integer ComId;

          private String Cescription;
        private String name;




    @OneToMany(mappedBy = "Com", cascade = CascadeType.ALL)
    @ToString.Exclude
    private List<SubCom> subCom = new ArrayList<>(); // *** Has a refrence


    //**** Constructor  

       public Com( Integer ComId , String Cescription,  String name) {
        this.ComId=ComId;
        this.Cescription = Cescription;
        this.name = name;

    }


}

我的第二个pojo是

@Entity
@Table(name = "sub_com")
@Data
public class SubCom implements Serializable {

  @Id
  @GeneratedValue(strategy = GenerationType.IDENTITY)
  @Column(name = "sub_com_id")
  private Integer Id;

  private String name;
  private String Cescription;


  @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.MERGE)
  @JoinColumn(name = "com_Id")
  private Com com; //****** Refernce to above table.

  }

我已经执行了如下所示的本地sql查询

SELECT
    c.com_Id,
    c.Cescription,
    sc.sub_domain_id
    FROM Com AS c
         INNER JOIN SubCom AS sc ON c.com_Id  = sc.com_Id

Answer 1

您必须选择所有必要的列以构造两个实体。

比这个例子中的映射更像：

@NamedNativeQuery(
    name = "find_person_with_phones_by_name",
    query =
        "SELECT " +
        "   pr.id AS \"pr.id\", " +
        "   pr.name AS \"pr.name\", " +
        "   pr.nickName AS \"pr.nickName\", " +
        "   pr.address AS \"pr.address\", " +
        "   pr.createdOn AS \"pr.createdOn\", " +
        "   pr.version AS \"pr.version\", " +
        "   ph.id AS \"ph.id\", " +
        "   ph.person_id AS \"ph.person_id\", " +
        "   ph.phone_number AS \"ph.number\", " +
        "   ph.phone_type AS \"ph.type\" " +
        "FROM Person pr " +
        "JOIN Phone ph ON pr.id = ph.person_id " +
        "WHERE pr.name LIKE :name",
    resultSetMapping = "person_with_phones"
)
 @SqlResultSetMapping(
     name = "person_with_phones",
     entities = {
         @EntityResult(
             entityClass = Person.class,
             fields = {
                 @FieldResult( name = "id", column = "pr.id" ),
                 @FieldResult( name = "name", column = "pr.name" ),
                 @FieldResult( name = "nickName", column = "pr.nickName" ),
                 @FieldResult( name = "address", column = "pr.address" ),
                 @FieldResult( name = "createdOn", column = "pr.createdOn" ),
                 @FieldResult( name = "version", column = "pr.version" ),
             }
         ),
         @EntityResult(
             entityClass = Phone.class,
             fields = {
                 @FieldResult( name = "id", column = "ph.id" ),
                 @FieldResult( name = "person", column = "ph.person_id" ),
                 @FieldResult( name = "number", column = "ph.number" ),
                 @FieldResult( name = "type", column = "ph.type" ),
             }
         )
     }
 ),

请在Hibernate文档中找到更多信息：https://docs.jboss.org/hibernate/orm/5.4/userguide/html_single/Hibernate_User_Guide.html

通过联接获取第二个实体的记录

1 个答案: