Spring Data JPA中使用规范进行内部连接和分组

Question

我正在尝试获取员工的详细信息，其雇员类型是职员，其加入日期是最近的一个。

SQL Server 2008中的查询如下所示：

select 
 * 
from 
  employee jj 
  inner join
  (
    select 
      max(join_date) as jdate,
      empltype as emptype 
    from 
      employee 
    where 
      empltype='clerk' 
    group by empltype
  ) mm 
    on jj.join_date=mm.jdate and jj.empltype=mm.emptype;

int seqid;(sequence id)
String empltype:
Date joindate;
String role;

Predicate toPredicate(Root<employee> root,CriteriaQuery <?> query,CriteriaBuilder cb)
{
            Predicate pred=null;
            // Returning the rows matching the joining date
            pred=cb.equal(root<Emplyoee_>.get("joindate"));
            //**//

}

提前致谢

Answer 1

我在记事本上写了这个，它还没有经过测试，但我想你在找类似的东西

QEmployee e1 = new QEmployee("e1");
QEmployee e2 = new QEmployee("e2");

PathBuilder<Object[]> eAlias = new PathBuilder<Object[]>(Object[].class, "eAlias");
JPASubQuery subQuery = JPASubQuery().from(e2)
                                    .groupBy(e2.empltype)
                                    .where(e2.empltype.eq('clerk'))
                                    .list(e2.join_date.max().as("jdate"), e2.emptype)

jpaQuery.from(e1)
        .innerJoin(subQuery, eAlias)
        .on(e1.join_date.eq(eAlias.get("jdate")), e1.emptype.eq(eAlias.get("emptype")))
        .list(qEmployee);