Question

我有两张桌子＆＃34; table1＆＃34;和＆＃34; table2＆＃34;。对于table1中的每一行，table2中可以有多行。为了获取数据，我创建了以下类

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SharedPreferences.Editor edit = prefs.edit();
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edit.commit();

和

@Entity
@EntityListeners(AuditingEntityListener.class) 
@Table(name = "table1", schema = "dbo")
public class Table1 extends BaseDomain{
  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  private Integer vendorId;
  private String name;
  private Integer companyId;
  private String details;

  @OneToMany(mappedBy = "table1",cascade = CascadeType.ALL)
  @NotFound(action = NotFoundAction.IGNORE)
  private Set<Table2> table2;

 // getter and setter for above
}

然后我创建了像这样的存储库

    @Entity
    @EntityListeners(AuditingEntityListener.class) 
    @Table(name = "table2", schema = "dbo")
    public class Table2 extends BaseDomain{
      @Id
      @GeneratedValue(strategy = GenerationType.AUTO)
      private Integer id;
      private Integer cityId;
      private Integer companyId;
      @OneToOne
      @JoinColumn(name = "vendorId")
      @JsonIgnore
      private Table1 table1;

     // getter and setter for above
    }

这将根据companyId给出table1中的行列表，并给出表2中具有相应vendorId的行列表。

直到现在我的实施工作正常。

回复是

@Transactional
public interface Table1Repository  extends JpaRepository<Table1, Integer> {

  Page<Table1> findByCompanyId(Integer companyId,Pageable pageable);

}

目前我正在将companyId作为输入并给出上述回复。现在我想把cityId作为输入。因此table2输出应仅包含给定cityId的详细信息。

所以任何人都可以帮助我如何实现它？我不想为table1和table2创建另一个模型。

Answer 1

Page<Table1> findByCompanyIdAndTable2_CityId(Integer companyId, Integer cityId, Pageable pageable);

此方法将负责cityId的提取。下划线（_）指嵌套字段。但要实现这一点，您需要急切地获取Table2。

@OneToMany(mappedBy = "table1",cascade = CascadeType.ALL, fetch=FetchType.EAGER)

JPA中的条件where子句

1 个答案: