我如何将此 mysql 语句转换为 hql答案

【问题标题】：How i can convert this mysql statement to hql我如何将此 mysql 语句转换为 hql
【发布时间】：2020-08-31 17:12:14
【问题描述】：

我如何将这个 mysql 语句转换为 hql

    select invisit, notinVisit
    FROM(
    select COUNT(*) as invisit  from consultations cons join patients p 
    on cons.patient_id = p.id
    where cons.datefin > now()) as A
    JOIN 
    ( select COUNT(*) as notinVisit from consultations cons join patients p 
    on cons.patient_id = p.id
    where cons.datefin < now()) as B

我的尝试是

"select com.organ.project.service.DTO.PatientsByInConsultationDTO(invisit, notinVisit)\n" +
            "FROM(\n" +
            "select COUNT(*) as invisit  from Consultation cons join Patient p \n" +
            "on cons.patient.id = p.id\n" +
            "where cons.datefin > now()) as A\n" +
            "JOIN \n" +
            "( select COUNT(*) as notinVisit from Consultation cons join Patient p \n" +
            "on cons.patient.id = p.id\n" +
            "where cons.datefin < now()) as B\n"

但我收到了这个错误：

 ...org.hibernate.hql.internal.ast.QuerySyntaxException: unexpected token: ( near line 2, column 5...

原来hql不支持子查询

ps： 我需要 invisit、notinVisit 属性作为构造函数参数来返回包含结果的对象

【问题讨论】：

标签： mysql hibernate jpa spring-data-jpa hql

【解决方案1】：

尝试使用这个

select new com.organ.project.service.DTO.PatientsByInConsultationDTO(
       sum(case when cons.datefin > now() then 1 else 0 end) as invisit,
       sum(case when cons.datefin < now() then 1 else 0 end) as notinVisit) 
from Consultation cons join Patient p on cons.patient.id = p.id
group by cons.id

【讨论】：