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Hibernate 如何在jpa中检索记录和字段值_Hibernate_Jpa - Fatal编程技术网

Hibernate 如何在jpa中检索记录和字段值

Hibernate 如何在jpa中检索记录和字段值,hibernate,jpa,Hibernate,Jpa,考虑以下实体: @Entity public class User { private int id; private String name; private String family; //getters and setters } 和sql: select * from (select row_.*, first_Value(rownum) over (order by rownum desc) countـ from (select *

考虑以下实体:

@Entity
public class User {
    private int id;
    private String name;
    private String family;
    //getters and setters
}
和sql:

select * 
from (select row_.*, first_Value(rownum) over (order by rownum desc) countـ
      from (select * from user) row_)
正如您可能猜到的,sql的结果集由四列组成:

id, name, family, countـ
我通过以下方式检索用户表记录作为用户列表:

String myQuery = "select * from (select row_.*, first_Value(rownum) over (order by rownum desc) count_
      from (select * from user) row_)"
Query query = entityManager.createNativeQuery(myQuery, User.class);
List<User> users = query.getResultList();
现在的问题是,我如何通过调用getResultList或任何其他hibernate等价物一次来检索用户列表并计算值

我使用Oracle 11g和Jpa Hibernate实现3.6


提前谢谢。

我想count是个保留字。我会像这样重新组织您的查询:

String myQuery = "SELECT "
        + "id, "
        + "name, "
        + "family, "
        + "cnt "
        + "FROM ("
            + "SELECT "
            + "user.*, "
            + "first_Value(rownum) over (order by rownum desc) AS cnt "
            + "FROM "
            + "user "
        + ")";
    Query query = entityManager.createNativeQuery(myQuery);
    List<Object[]> resultList = query.getResultList();

当然,您也可以使用*而不是id、姓名、家族、cnt。

代码中的生成器是什么?对不起,我用myQuery替换了它。。。