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

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@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();

@Datapublic class UserMorningReport {    private String userName;    private String plan;}

@Slf4jpublic 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

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