我在我的项目中使用JPA。

我遇到一个查询,我需要对五个表进行连接操作。因此,我创建了一个返回五个字段的本机查询。

现在我想将结果对象转换为java POJO类,其中包含相同的五个字符串。

在JPA中有任何方法可以直接将结果转换为POJO对象列表吗??

我想出了如下的解决办法。

@NamedNativeQueries({  
    @NamedNativeQuery(  
        name = "nativeSQL",  
        query = "SELECT * FROM Actors",  
        resultClass = db.Actor.class),  
    @NamedNativeQuery(  
        name = "nativeSQL2",  
        query = "SELECT COUNT(*) FROM Actors",  
        resultClass = XXXXX) // <--------------- problem  
})  

现在在resultClass中,我们需要提供一个实际的JPA实体类吗? 或 我们可以将其转换为包含相同列名的任何JAVA POJO类?


当前回答

使用ResultSet的旧样式

@Transactional(readOnly=true)
public void accessUser() {
    EntityManager em = this.getEntityManager();
    org.hibernate.Session session = em.unwrap(org.hibernate.Session.class);
    session.doWork(new Work() {
        @Override
        public void execute(Connection con) throws SQLException {
            try (PreparedStatement stmt = con.prepareStatement(
                    "SELECT u.username, u.name, u.email, 'blabla' as passe, login_type as loginType FROM users u")) {
                ResultSet rs = stmt.executeQuery();
                ResultSetMetaData rsmd = rs.getMetaData();
                for (int i = 1; i <= rsmd.getColumnCount(); i++) {
                    System.out.print(rsmd.getColumnName(i) + " (" + rsmd.getColumnTypeName(i) + ") / ");
                }
                System.out.println("");
                while (rs.next()) {
                    System.out.println("Found username " + rs.getString("USERNAME") + " name " + rs.getString("NAME") + " email " + rs.getString("EMAIL") + " passe " + rs.getString("PASSE") + " email " + rs.getInt("LOGIN_TYPE"));
                }
            }
        }
    });
}

其他回答

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。

不确定这是否适合这里,但我有类似的问题,并为我找到了以下简单的解决方案/例子:

private EntityManager entityManager;
...
    final String sql = " SELECT * FROM STORE "; // select from the table STORE
    final Query sqlQuery = entityManager.createNativeQuery(sql, Store.class);

    @SuppressWarnings("unchecked")
    List<Store> results = (List<Store>) sqlQuery.getResultList();

在我的情况下,我必须使用在其他地方的字符串中定义的SQL部分,所以我不能只使用NamedNativeQuery。

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();
}

既然其他人已经提到了所有可能的解决方案,我将分享我的变通方案。

在我使用Postgres 9.4的情况下,当我和Jackson一起工作时,

//Convert it to named native query.
List<String> list = em.createNativeQuery("select cast(array_to_json(array_agg(row_to_json(a))) as text) from myschema.actors a")
                   .getResultList();

List<ActorProxy> map = new ObjectMapper().readValue(list.get(0), new TypeReference<List<ActorProxy>>() {});

我相信您可以在其他数据库中找到相同的方法。

另供参考,JPA 2.0本机查询结果为映射