组上的java JPA标准API空检查

Question

我需要返回结果数组，其中分组列上的数据可能包含空值，当前跳过这些值，同时我还希望将它们分组

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

它适用于具有内部状态的实体！=null，但是可以在没有connected InternalStatus的情况下对其进行分组，然后我希望按null对其进行分组

当前示例结果：

// ... 
        {
            "username": "admin@gmail.com",
            "internalStatus": "Do stuff",
            "numberOfDoc1": 0,
            "numberOfDoc2": 2,
            "numberOfDoc3": 1,
            "numberOfDoc4": 0,
            "sumOfDoc": 3
        },

我希望它也能产生这样的结果：

        {
            "username": "admin@gmail.com",
            "internalStatus": null,
            "numberOfDoc1": 4,
            "numberOfDoc2": 1,
            "numberOfDoc3": 5,
            "numberOfDoc4": 0,
            "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

有没有办法将自动创建的交叉联接更改为考虑null

Answer 1

试试这个

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