JPA Criteria API-对GROUP BY进行空检查

问题描述

我需要返回结果数组，其中分组列上的数据可能包含空值，并且当前将跳过这些值，而我也希望将它们分组。

我的实体：

public class UserEntity {
// ...
    @Basic
    @Column(name = "username")
    private String username;
}

public class ZgloszenieEntity {
// ... 

    @ManyToOne
    @JoinColumn(name = "assigned_user_id" )
    @OrderBy("username")
    private UserEntity assignedUser;

    @ManyToOne(targetEntity = InternalStatusEntity.class)
    @JoinColumn(name = "internal_status")
    @NotAudited
    private InternalStatusEntity internalStatus;
}

受影响的代码：

  CriteriaBuilder cb = em.getCriteriaBuilder();
        CriteriaQuery<EfficiencyStatusReportDTO> cq = cb.createquery(EfficiencyStatusReportDTO.class);
        Root<ZgloszenieEntity> root = cq.from(ZgloszenieEntity.class);
        Join<UserEntity,ZgloszenieEntity> join = root.join("assignedUser");

        cq.multiselect(join.get("username"),root.get("internalStatus").get("description"),getNumberOfDocOfType(cb,root,"X-1"),getNumberOfDOcOfType(cb,"X-2"),"X-3"),"X-4"),cb.count(root)
        );
        cq.groupBy(join.get("username"),root.get("internalStatus").get("description"));

对于具有internalStatus！= null的实体来说，它可以很好地工作，但是有可能在没有连接InternalStatus的情况下拥有它，我希望将其按null分组。

当前示例结果：

// ... 
        {
            "username": "[e-mail]","internalStatus": "Do stuff","numberOfDoc1": 0,"numberOfDoc2": 2,"numberOfDoc3": 1,"numberOfDoc4": 0,"sumOfDoc": 3
        },

我也希望得到这样的结果：

        {
            "username": "[e-mail]","internalStatus": null,"numberOfDoc1": 4,"numberOfDoc2": 1,"numberOfDoc3": 5,"sumOfDoc": 10
        },

JPA生成的查询：

select userentity1_.username                                          as col_0_0_,internalst2_.description                                       as col_1_0_,count(case when zgloszenie0_.form_type='X-1' then 1 else null end) as col_2_0_,count(case when zgloszenie0_.form_type='X-2' then 1 else null end) as col_3_0_,count(case when zgloszenie0_.form_type='X-3' then 1 else null end) as col_4_0_,count(case when zgloszenie0_.form_type='X-4' then 1 else null end) as col_5_0_,count(zgloszenie0_.uid)                                        as col_6_0_
from zgloszenie zgloszenie0_
         inner join user userentity1_ on zgloszenie0_.assigned_user_id = userentity1_.id
         cross join internal_status internalst2_
where zgloszenie0_.internal_status = internalst2_.id
  and 1 = 1
group by userentity1_.username,internalst2_.description
order by userentity1_.username desc

是否可以将自动创建的交叉联接更改为考虑为空？

解决方法

尝试一下

 CriteriaBuilder cb = em.getCriteriaBuilder();
 CriteriaQuery<EfficiencyStatusReportDTO> cq = 
     cb.createQuery(EfficiencyStatusReportDTO.class);
 Root<ZgloszenieEntity> root = cq.from(ZgloszenieEntity.class);
 Join<ZgloszenieEntity,UserEntity> assignedUser = root.join("assignedUser",JoinType.LEFT);
 Join<ZgloszenieEntity,InternalStatus> internalStatus = root.join("internalStatus",JoinType.LEFT);

 cq.multiselect(assignedUser.get("username"),internalStatus.get("description"),getNumberOfDocOfType(cb,root,"X-1"),getNumberOfDOcOfType(cb,"X-2"),"X-3"),"X-4"),cb.count(root)
 );
 cq.groupBy(assignedUser.get("username"),internalStatus.get("description"));