JPA 查询等于两个属性之间 [英] JPA query equals between two properties

问题描述

是否可以使用 JPA 自定义查询来比较两个属性?

It is possible to compare two properties with JPA custom query ?

注册.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;

}

Rate.java :

Rate.java :

@Entity
public class Rate implements Serializable {

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

@ManyToOne
private Activity activity;
}

规划.java :

@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;

我只会检索 rate.activity.id =planning.activity.id 类似的东西:

I would retrieve only rate.activity.id = planning.activity.id with something like :

public interface RegistrationDao extends CrudRepository<Registration, UUID> {

    findByPlanningAndRateActivityEqualsPlanningActivityOrderByEntryLastnameAsc(Planning planning);

 }

谢谢

推荐答案

与其使用很长的方法名称，不如使用 JPA 查询，它会是这样的.希望这会有所帮助.

Rather than having a long method name you can have a JPA Query it will be something like this. Hope this helps.

@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);

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