Hibernate：如何通过连接字符串来创建 JPQL？

Question

我正在尝试寻找一种连接字符串以创建 JPA 原生查询的方法。目前我有 3 个查询，只是因为有 1 个或 2 个不同的单词。例如：

String queryAsc = select emp.id,role_id from employee left join roles 
                         on roles.id = emp.role_id      
                         where emp.id = :id
                         order by employee.created_date asc;

String queryDesc = select emp.id,role_id from employee left join roles 
                         on roles.id = emp.role_id
                         where emp.id = :id
                         order by employee.created_date desc;

仅仅因为“asc”和“desc”这个词的不同，我不得不用JPA查询写了2个不同的repository方法：

@Query(value = queryAsc,
       countQuery = countToMyEmployees, nativeQuery = true)
Page<Object[]>  findMyEmployeesAsc(@Param("id") String id, Pageable pageable);

 
@Query(value = queryDesc,
       countQuery = countToMyEmployees, nativeQuery = true)
Page<Object[]>  findMyEmployeesDesc(@Param("id") String id, Pageable pageable);

有没有办法在 JPA 查询中连接两个字符串以避免这种重复？

Answer 1

尝试使用调用方的排序。

 if (condition) {
    obj.findMyEmployees("123",pageableObject , new Sort("employee.created_date "));
    } else {
    obj.findMyEmployees("234",pageableObject ,new Sort(Sort.Direction.DESC,"employee.created_date"));
    }

并重写查询如下：

String query= select emp.id,role_id from employee left join roles 
                         on roles.id = emp.role_id
                         where emp.id = :id;
@Query(value = query,
       countQuery = countToMyEmployees, nativeQuery = true)
Page<Object[]>  findMyEmployees(@Param("id") String id, Pageable pageable, Sort sort);