JPA 2 - 获取@OneToMany 关系中列表的最后一个元素

Question

我有一个实体，其实体名为 Deadline，与名为 DeadlineDate 的实体具有@OneToMany 关系。

截止日期：

@Entity
@Table(name="deadline")
@Cache(alwaysRefresh=true)
public class Deadline implements Serializable { 
    @Id
    @GeneratedValue(strategy= GenerationType.AUTO)
    @Column(name = "id", nullable = false)
    private Long id;


    @OneToMany(mappedBy = "deadline", cascade = PERSIST)
    @OrderBy("id ASC")
    private List<DeadlineDate> datesList = new ArrayList<DeadlineDate>();

    //...
}

DeadlineDate 实体：

@Entity
@Table(name="deadlinedate")
public class DeadlineDate implements Serializable {     
    @Id
    @GeneratedValue(strategy = IDENTITY)
    @Column(name = "id")
    private Long id;

    @Basic
    @Column(name = "date", nullable = false)
    @Temporal(DATE)
    private Date date;

    @ManyToOne
    @JoinColumn(nullable = false, name = "deadline_id")
    private Deadline deadline;
    //...
}

getAllDeadlines 方法：

private List<Deadline> getAllDeadlines(){
        CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
        CriteriaQuery<Deadline> query = criteriaBuilder.createQuery(Deadline.class);
        Root<Deadline> root = query.from(Deadline.class);

        Join<Deadline, DeadlineDate> joinDeadDate = root.join("datesList");

        List<Predicate> predicateList = new ArrayList<Predicate>();             

        Date date = new Date();

        //other unrelated predicates

        //i'm stuck here - begin
        predicateList.add( criteriaBuilder.lessThan(joinDeadDate.<predicateArray>get("date"), date));       
        //i'm stuck here - end

        Predicate[] predicateArray = new Predicate[predicateList.size()];
        predicateList.toArray(predicateArray);
        query.where(predicateArray);        

        TypedQuery<Deadline> typedQuery = em.createQuery(query);
        return typedQuery.getResultList();
    }

我想查询已超过当前系统日期的截止日期。截止日期具有存储用户所做更改的历史记录（到期日期可以向前或向后修改）。当前截止日期被认为是截止日期表中最大的 id。我正在尝试使用 Criteria api，但不知道如何使其仅返回 datesList 的最后一个值，以便在 criteriaBuilder 的 lessThan 方法中使用。使用当前代码，我获得了特定截止日期的所有截止日期记录。

提前致谢！

Answer 1

如果您只需要为每个 Deadline 使用具有最大 id 的 DeadlineDate，则需要一个子查询（未测试）：

Subquery<Long> sq = query.subquery(Long.class);
Root<DeadlineDate> deadlineDate = query.from(DeadlineDate.class);
Root<DeadlineDate> deadlineDateSQ = sq.from(DeadlineDate.class);
Path<Deadline> deadline = deadlineDate.get("deadline");
Path<Deadline> deadlineSQ = deadlineDateSQ.get("deadline");
sq.select(criteriaBuilder.max(deadlineSQ.<Long>get("id")));
sq.groupBy(deadlineSQ);
... //create the list of other predicates
sq.where(predicateListSQ);
query.select(deadline);
Predicate predicate1 = criteriaBuilder.in(deadlineDate.get("id")).value(sq);
Predicate predicate2 = criteriaBuilder.lessThan(deadlineDate.<Date>get("date"), date);
query.where(criteriaBuilder.and(predicate1, predicate2));

另请参阅：

• Compare Date entities in JPA Criteria API
• JPA 2.0, Criteria API, Subqueries, In Expressions

Answer 2

使用setMaxResults 限制返回的记录数。对输出进行排序，以便您想要的记录是第一个。

typedQuery.setMaxResults(1);