Spring data jpa 自定义查询返回，用FastJson把Map转换为JavaBean

@Query("select new map(u.userName as userName , m.plan  as plan ) from User u , MorningReport m where u.id = '1' and  m.id = 1 ")
public Map<String,String> testMap();

@Data
public class UserMorningReport {
private String userName;
private String plan;
}

@Slf4j
public class CommentTest extends WorkTileApplicationTests{

@Autowired
private UserDao userdao;

@Test
public void  testUser(){

//自定义查询用
Map<String, String> testMap = userdao.testMap();
String itemJSONObj = JSON.toJSONString(testMap);
System.out.println("itemJSONObj=="+itemJSONObj);
UserMorningReport group = JSON.parseObject(itemJSONObj, UserMorningReport.class);
System.out.println(group.getPlan());
System.out.println(group.getUserName());
log.info("group={}", group);
}
}

与select new map相似的还有select new list、select new set。只研究了此一个，没有研究别的，但大致原理应该是差不多的。

参考—– Hibernate调优之select new map()

http://blog.csdn.net/z69183787/article/details/41362093