使用spring jpa 时,利用nativeQuery,获取数据,无需新建实体,按照别名返回Json数据
刚开始是这样写的
@Query(value = "SELECT ll.user_id id ,u.catong_img catong_img,ll.locationId location_id,ll.address address,ll.latitude latitude,ll.longitude longitude,ll.update_time update_time,ll.user_id user_id FROM t_user as u left JOIN (SELECT l.id locationId, address,latitude,longitude,l.update_time,l.user_id FROM t_user_location AS l RIGHT JOIN ( SELECT user_id, max( update_time ) update_time FROM t_user_location GROUP BY user_id ) AS t ON l.user_id = t.user_id AND l.update_time = t.update_time ) ll ON u.zu_id =?1 and u.id = ll.user_id",nativeQuery = true)
@Transactional
List getAllUserLocationByZuId(long zuId);
发现返回的数据是这样的,设置的别名没有返回
[
[
1,
"吕志豪",
"",
3,
"3",
3,
3,
1513670259000,
1
],
[
2,
"胡勇",
"",
2,
"2",
2,
2,
1513151843000,
2
]
]
之后是这样写的,使用EntityManager
@PersistenceContext
private EntityManager em;
@Override
public List getAllUserLocation(long zuId) {
return em.createNativeQuery("SELECT u.name name ,u.catong_img catongImg,ll.locationId locationId,ll.address address,ll.latitude latitude,ll.longitude longitude,ll.update_time updateTime,ll.user_id userId FROM t_user as u left JOIN (SELECT l.id locationId, address,latitude,longitude,l.update_time,l.user_id FROM t_user_location AS l RIGHT JOIN ( SELECT user_id, max( update_time ) update_time FROM t_user_location GROUP BY user_id ) AS t ON l.user_id = t.user_id AND l.update_time = t.update_time ) ll ON u.zu_id =1 and u.id = ll.user_id")
.unwrap(SQLQuery.class)
.setResultTransformer(
AliasToEntityMapResultTransformer.INSTANCE
)
.list();
}
结果返回
[
{
"catongImg": "",
"address": "3",
"locationId": 3,
"latitude": 3,
"name": "吕志豪",
"updateTime": 1513670259000,
"userId": 1,
"longitude": 3
},
{
"catongImg": "",
"address": "2",
"locationId": 2,
"latitude": 2,
"name": "胡勇",
"updateTime": 1513151843000,
"userId": 2,
"longitude": 2
}
]
原文地址:https://www.jianshu.com/p/e6bc577d03f0