Question

我的对象：用户和凭据 - 多对多关系，但用户有一个参数

我希望所有用户都能获得循环中每个凭据的某个参数

要求：必须批量加载用户。

简单吧？

所以我有3张桌子：

@Table(name = "CRED")
public class Credential {

    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    @Column(name="CRED_ID")     
    Long credentialId;

    @OneToMany(fetch=FetchType.LAZY, cascade = CascadeType.ALL, mappedBy = "credential")
    @BatchSize(size = Search.DEFAULT_PAGE_SIZE)
    private Set<UserCredentialMapping> userCredentialMappingSet;  
}

@Table(name = "USER_CRED")
public class UserCredentialMapping {

    @JoinColumn(name = "user_id", referencedColumnName = "user_id")
    @ManyToOne
    @Filter(name="paramFilter", condition="param = :param")
    private User user;

    @JoinColumn(name = "cred_id", referencedColumnName = "cred_id")
    @ManyToOne
    private Credential credential;
}

@Table(name = "USER")
public class User  {

    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    @Column(name="USER_ID")     
    Long userId;

    @Column(name = "PARAM")
    String param
}

我在一个地方进行查询并返回结果：

    String hqlQuery =   "select c from UserCredentialMapping m " +
        " inner join m.credential c" +
        " inner join m.user u" +
        " where u.param = :param" +
        " and c.user_id in (:users)" ;

        Session session = getSession();
        //desparetly trying to set filter
        session.enableFilter("paramFilter").setParameter("param", param);

    Query query = session.createQuery(hqlQuery);
    query.setParameterList("users", USERLIST);
    query.setParameter("param", someparam);

    List<Credential> credentialList = (List<Credential>)query.list();
    return credentialList;

平均时间对每个凭证进行一些处理，现在我需要获得给定参数的用户列表：

    for(Credential credential : credentialList){

        //following line makes hibernate query for users
        Iterator<CredentialMapping> mappingIterator = e.getUserCredentialMappingSet().iterator();

        while (mappingIterator.hasNext()){
            UserCredentialMapping userCred = mappingIterator.next();

            User user = userCred.getUser(); 
            DOEVILSTUFFTOINNOCENT(user);
    }

我的问题是迭代器生成SQL查询，该查询获取所有用户的凭证，而不是所有用户指定的参数用于凭证（换句话说，过滤器未被应用）

有任何建议如何让它发挥作用？

谢谢！

Answer 1

我通过向Credential类添加Many To Many映射来解决它：

@ManyToMany(fetch = FetchType.LAZY)
@NotFound(action = NotFoundAction.IGNORE)
@JoinTable(name = "USER_CRED", 
    joinColumns = { 
        @JoinColumn(name = "CRED_ID") }, 
    inverseJoinColumns = { 
        @JoinColumn(name = "USER_ID") })
@Filter(name="param", condition="PARAM = :param")
@BatchSize(size = Search.DEFAULT_PAGE_SIZE)
private Set<User> users;

我无法通过UserCredential Mapping ...

来完成它

过滤延迟初始化集合

1 个答案: