Question

@Entity
@Table(name="User")
public class User implements Serializable{
    private String username; //Key

    // Getters and setters
}    

@Entity
@Table(name="POITable")
public class POI implements Serializable{
    private Integer id; //Key
    private String name;

    // Getters and setters
}

@Entity
@Table(name="VisitedPlaceTable")
public class VisitedPlace implements Serializable{
    private Integer id; //Key
    private User user; //Many TO One relation
    private POI poi; //Many TO One relation

    // Getters and setters
}

POITable

===================
| id |    name    |
ــــــــــــ+ـــــ+
| 1  | poi name 1 |
-----+------------+
| 2  | poi name 2 |
===================

VisitedPlaceTable

=================================
| id |  user_username  | poi_id |
ــــــــ+ـــــــــــــــــ+ـــــ+
| 3  | xxxx@gmail.com  |   1    |
-----+-----------------+--------+
| 4  | xxxx@gmail.com  |   2    |
=================================

此原生查询

SELECT CONCAT('"',model.user_username,'","',**ARRAY_TO_STRING(ARRAY(SELECT DISTINCT visitedPlace.poi_id FROM VisitedPlaceTable visitedPlace WHERE visitedPlace.user_username = model.user_username),'|')**,'"','') FROM TourismPOIVisited model  INNER join POITable poi on poi.id = model.poi_id WHERE model.user_username='xxxx@gmail.com' GROUP BY model.user_username

返回下一个结果（TRUE和EXPECTED结果:)）

“xxxx@gmail.com”， “1 | 2”

HQL中

但

SELECT CONCAT('\"',model.user.username,'\",\"',ARRAY_TO_STRING(ARRAY(SELECT DISTINCT visitedPlace.poi.id FROM VisitedPlace visitedPlace WHERE visitedPlace.user.username = model.user.username),'|'),'\"','\n') FROM VisitedPlace model WHERE model.user.username='xxxx@gmail.com' GROUP BY model.user.username

它开火

第1:70行：意外令牌： SELECT

那么如何使用 ARRAY_TO_STRING（ARRAY（））或任何其他等效函数在HQL中执行此操作？

Hibernate：如何在HQL中使用ARRAY_TO_STRING（ARRAY（））

0 个答案: