Question

可以将两个属性与JPA自定义查询进行比较吗？

Registration.java：

=if(or(isna(index('SSI Tracker'!$A:$F,small(match(F25,'SSI Tracker'!$D:$D,0),2),2)),index('SSI Tracker'!$A:$F,small(match(F25,'SSI Tracker'!$D:$D,0),2),2)=""),"",concatenate(index('SSI Tracker'!$A:$F,small(match(F25,'SSI Tracker'!$D:$D,0),2),1)," / ",index('SSI Tracker'!$A:$F,small(match(F25,'SSI Tracker'!$D:$D,0),2),2)))

Rate.java：

@Entity
public class Registration implements Serializable {

@Id
@GeneratedValue(generator = "uuid")
private UUID uuid;

@ManyToOne(targetEntity=Entry.class)
private Entry entry;

@ManyToMany(targetEntity=Rate.class,fetch = FetchType.EAGER)
private List<Rate> rate;

@ManyToMany(fetch = FetchType.EAGER)
private List<Planning> planning;

}

Planning.java：

@Entity
public class Rate implements Serializable {

@Id
@GeneratedValue(generator = "uuid")
private UUID uuid;

@ManyToOne
private Activity activity;
}

我只会检索rate.activity.id = planning.activity.id，例如：

@Entity
public class Planning implements Serializable {

@Id
@GeneratedValue(generator = "uuid")
private UUID uuid;

@OneToOne
private Location location;

@OneToOne(fetch = FetchType.EAGER)
private Coach coach;

@OneToOne
private Activity activity;

由于

Answer 1

您可以拥有一个JPA查询，而不是拥有一个长方法名称。希望这会有所帮助。

@Query("select * FROM Registration REG,Entry E, Rate R,Planning P WHERE REG.id = :id and R.activity=P.activity ORDER BY E.lastname")

public List<Registration> <MethodName>(@Param("id") String registrationid);

Answer 2

我已针对此解决方案进行了更改，但我认为这不是一个好方法（其中rateActivity和planningActivity必须具有相同的ID）

public interface RegistrationDao extends CrudRepository<Registration, UUID> {

  Iterable<Registration> findByPlanningAndRateActivityEqualsAndPlanningActivityEqualsOrderByEntryLastnameAsc(Planning planning,Activity rateActivity,Activity planningActivity);

}

JPA查询等于两个属性

2 个答案: