1
我做JPQL查询,像这样自定义选择@Query JSON键
@Repository
@Transactional
public interface UserFlightDao extends CrudRepository<UserFlight, Long> {
@Query("SELECT uf.departureGps, uf.flight.id, uf.flight.flightNumber, uf.flight.airline.name, uf.flight.departureDate, " +
"uf.flight.departureAirport.name FROM UserFlight uf WHERE user.id=?1")
List<UserFlight> getUserFlights(Long userId);
}
UserFlight
包含Flight
对象,我选择从两个UserFlight
和Flight
对象值,并将其返回给用户的JSON 。
首先,我认为使用List<UserFlight>
作为返回类型(即使它有效)是错误的,因为技术上我没有返回完整的UserFlight
对象。对?也许我应该切换到List<Object>
。
第二件事,我希望返回给用户的json包含关键名称。目前我得到的json包含了一组对象,没有它们各自的键名。响应示例:
[
[
"sdf",
1,
234234,
"American Airline",
{
"dayOfMonth": 13,
"dayOfWeek": "TUESDAY",
"dayOfYear": 286,
"monthValue": 10,
"month": "OCTOBER",
"year": 2015,
"hour": 18,
"minute": 41,
"nano": 0,
"second": 39,
"chronology": {
"id": "ISO",
"calendarType": "iso8601"
}
},
"dummy airport"
],
[
"asfsaf",
1,
234234,
"American Airline",
{
"dayOfMonth": 13,
"dayOfWeek": "TUESDAY",
"dayOfYear": 286,
"monthValue": 10,
"month": "OCTOBER",
"year": 2015,
"hour": 18,
"minute": 41,
"nano": 0,
"second": 39,
"chronology": {
"id": "ISO",
"calendarType": "iso8601"
}
},
"dummy airport"
]
]
任何想法如何获取关键名称以及值?我应该在从存储库收到List<Object>
后手动构建json还是有一个更简单的方法?
这是我怎么称呼getUserFlights
感谢您的提示。我创建了一个类来保存你所建议的值(类没有注释,只有列子集的setter/getters),我使用了'SELECT new com.foo.bar.model.UserFlightLight(arg1,arg2..etc) '在jpql查询中,一切都很好。 – prettyvoid