我们通过以下方式解决了这个问题:

   //Add actual table name here in Query
    final String sqlQuery = "Select a.* from ACTORS a"
    // add your entity manager here 
    Query query = entityManager.createNativeQuery(sqlQuery,Actors.class);
    //List contains the mapped entity data.
    List<Actors> list = (List<Actors>) query.getResultList();

最简单的方法就是使用投影。它可以将查询结果直接映射到接口，并且比使用SqlResultSetMapping更容易实现。

一个例子如下所示:

@Repository
public interface PeopleRepository extends JpaRepository<People, Long> {

    @Query(value = "SELECT p.name AS name, COUNT(dp.people_id) AS count " +
        "FROM people p INNER JOIN dream_people dp " +
        "ON p.id = dp.people_id " +
        "WHERE p.user_id = :userId " +
        "GROUP BY dp.people_id " +
        "ORDER BY p.name", nativeQuery = true)
    List<PeopleDTO> findByPeopleAndCountByUserId(@Param("userId") Long userId);

    @Query(value = "SELECT p.name AS name, COUNT(dp.people_id) AS count " +
        "FROM people p INNER JOIN dream_people dp " +
        "ON p.id = dp.people_id " +
        "WHERE p.user_id = :userId " +
        "GROUP BY dp.people_id " +
        "ORDER BY p.name", nativeQuery = true)
    Page<PeopleDTO> findByPeopleAndCountByUserId(@Param("userId") Long userId, Pageable pageable);

}



// Interface to which result is projected
public interface PeopleDTO {

    String getName();

    Long getCount();

}

投影接口中的字段必须与该实体中的字段匹配。否则字段映射可能会中断。

同样，如果你使用SELECT表。列表示法总是定义与实体名称匹配的别名，如示例所示。

使用Hibernate:

@Transactional(readOnly=true)
public void accessUser() {
EntityManager em = repo.getEntityManager();
    org.hibernate.Session session = em.unwrap(org.hibernate.Session.class);
    org.hibernate.SQLQuery q = (org.hibernate.SQLQuery) session.createSQLQuery("SELECT u.username, u.name, u.email, 'blabla' as passe, login_type as loginType FROM users u").addScalar("username", StringType.INSTANCE).addScalar("name", StringType.INSTANCE).addScalar("email", StringType.INSTANCE).addScalar("passe", StringType.INSTANCE).addScalar("loginType", IntegerType.INSTANCE)
        .setResultTransformer(Transformers.aliasToBean(User2DTO.class));

    List<User2DTO> userList = q.list();
}

如果查询不是太复杂，您可以这样做。在我的情况下，我需要使用H2 FT_Search结果查询来进行另一个查询。

var ftSearchQuery = "SELECT * FROM FT_SEARCH(\'something\', 0, 0)";
List<Object[]> results = query.getResultList();
List<Model> models = new ArrayList<>();
for (Object[] result : results) {
    var newQuery = "SELECT * FROM " + (String) result[0];
    models.addAll(entityManager.createNativeQuery(newQuery, Model.class).getResultList());
  }

也许有更干净的方法。

如果你使用Spring，你可以使用org.springframework.jdbc.core.RowMapper

这里有一个例子:

public List query(String objectType, String namedQuery)
{
  String rowMapper = objectType + "RowMapper";
  // then by reflection you can instantiate and use. The RowMapper classes need to follow the naming specific convention to follow such implementation.
} 

在上面的回答中，我尝试了很多事情。SQLmapper不知道该把它放在哪里。非托管pojo只是一个问题。我尝试了各种方法，其中一个简单的方法就是像往常一样。我使用的是hibernate-jpa-2.1。

List<TestInfo> testInfoList = factory.createNativeQuery(QueryConstants.RUNNING_TEST_INFO_QUERY)
                    .getResultList();

唯一需要注意的是POJO具有与查询相同的成员变量名(均为小写)。显然，我甚至不需要像在JPQL中使用TypedQueries那样告诉目标类和查询。

TestInfo.class

@Setter
@Getter
@NoArgsConstructor
@ToString
public class TestInfo {

    private String emailid;
    private Long testcount;

    public TestInfo(String emailId, Long testCount) {
        super();
        this.emailid = emailId;
        this.testcount = testCount;
    }

}