0

我有三个实体。我需要构建 Criteria API，如果唯一用户超过 userCount 变量，我可以在其中选择项目。

@Entity
@Table(name = "client")
public class Client {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Getter @Setter private Long id;
}

@Entity
@Table(name = "session")
public class Session {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Getter @Setter private Long id;

    @ManyToOne
    @JoinColumn(name = "client_id", referencedColumnName = "id")
    @Getter @Setter private Client client;

    @ManyToOne
    @JoinColumn(name = "project_id", referencedColumnName = "id")
    @Getter @Setter private Project project;
}

@Entity
@Table(name = "project")
public class Project {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Getter @Setter private Long id;
}

我想选择所有项目，其中唯一用户> = userCount。我在 jpql 中构造查询

@Query("select p from Project p where (select distinct count(cli.id) from Client as cli 
join Session sess on sess.client = cli 
join Project as proj on proj = sess.project 
where proj.id = p.id) >= :userCount")

我写了标准：

        CriteriaBuilder cb = em.getCriteriaBuilder();
        CriteriaQuery<Project> cq = cb.createQuery(Project.class);
        Root<Project> projectRoot = cq.from(Project.class);

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

        Subquery<Long> sub = cq.subquery(Long.class);
        Root<Client> subRoot = sub.from(Client.class);

        Join<Client, Session> sessionClientJoin = subRoot.join("sessions");
        Join<Session, Project> sessionProjectJoin = sessionClientJoin.join("project");

        sub.select(cb.count(subRoot.get("id"))).distinct(true);
        sub.where(cb.equal(projectRoot.get("id"), sessionProjectJoin.get("id")));

        predicates.add(cb.greaterThanOrEqualTo(sub, DefaultParamsHolder.NUMBER_OF_USERS));
        cq.select(projectRoot);
        cq.where(predicates.toArray(new Predicate[0]));
        List<Project> project = em.createQuery(cq).getResultList();

这工作正常，但此标准需要客户端类中的会话。

@Entity
@Table(name = "client")
public class Client {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Getter @Setter private Long id;

    @OneToMany(mappedBy = "client", fetch = FetchType.LAZY)
    private List<Session> sessions;
}

这对我很不利。我需要在客户端类中构建没有会话的 Criteria API。我试过这个：

        CriteriaBuilder cb = em.getCriteriaBuilder();
        CriteriaQuery<Project> cq = cb.createQuery(Project.class);
        Root<Project> projectRoot = cq.from(Project.class);

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

        Subquery<Long> sub = cq.subquery(Long.class);
        Root<Client> subRoot = sub.from(Client.class);

        Join<Session, Client> sessionClientJoin = subRoot.join("client");
        Join<Session, Project> sessionProjectJoin = sessionClientJoin.join("project");

        sub.select(cb.count(subRoot.get("id"))).distinct(true);
        sub.where(cb.equal(projectRoot.get("id"), sessionProjectJoin.get("id")));

        predicates.add(cb.greaterThanOrEqualTo(sub, DefaultParamsHolder.NUMBER_OF_USERS));
        cq.select(projectRoot);
        cq.where(predicates.toArray(new Predicate[0]));
        List<Project> project = em.createQuery(cq).getResultList();

这行不通。

java.lang.IllegalArgumentException: Unable to locate Attribute  with the the given name [client] on this ManagedType [com.engage.domain.model.statisctic.Client]

如果没有客户端类中的会话，我怎么能做到这一点？

4

1 回答 1

0

你有

Root<Client> subRoot = sub.from(Client.class);
Join<Session, Client> sessionClientJoin = subRoot.join("client");

此连接表示“连接到绑定到客户端属性的客户端类实体”。 Client没有属性client- 这就是错误所说的。

也许您想加入与客户的会话

   Join<Client, Session> sessionClientJoin = subRoot.join("sessions");

我的建议是开始使用静态生成元模型。这将保证编译时属性的有效性和类型安全。

https://www.baeldung.com/hibernate-criteria-queries-metamodel

于 2019-10-09T08:43:52.687 回答