Question

我有三个实体。我需要构造Criteria API，如果唯一用户数超过userCount变量，我可以选择项目。

@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;
}

我要选择所有项目，其中唯一用户> = userCount。我在jpql中构建查询

@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();

这项工作正常，但是此标准要求在“客户”类中进行会话。

@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;
}

这对我不利。我需要在Client类中没有会话的情况下构造Criteria API。我尝试过：

        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();

这不起作用。

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

在没有Client类会话的情况下如何做到这一点？

Answer 1

你有

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

此联接显示“联接到与客户机属性绑定的客户机类实体”。 Client没有属性client-这就是错误的意思。

也许您想加入到客户端的会话

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

我的建议是开始使用静态遗传元模型。这将保证属性的有效性和编译时的类型安全性。

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

具有单向关系的条件API

1 个答案: