JPA/Hibernate：@ManyToOne和@OneToOne标记为 FetchType.LAZY 和 optional = false 的关系，而不是在 em.find（） 上懒惰地加载？

我有以下实体（仅显示相关映射）：

@Entity
@Table(name = "PQs")
public class PQ implements Serializable
{
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column
    private Integer id;

    @Column
    private String name;

    @ManyToOne(fetch = FetchType.LAZY)                 // lazy XToOne
    @JoinColumn(name = "user_id", referencedColumnName = "person_id")
    private User user;

    @OneToOne(mappedBy = "pq", fetch = FetchType.LAZY) // lazy XToOne
    private Group group;

    @OneToOne(mappedBy = "pq", fetch = FetchType.LAZY) // lazy XToOne
    private Tendering tendering;

    ...
}

请注意上面的注释：与其他实体有三种关系：@XToOne

用户（具有简单 ID 为 PK 的 SecurityIdentity 子类，由表示所有者的 PQ 引用）：

@Entity
@Table(name = "Users")
@DiscriminatorValue(value = "user")
public class User extends SecurityIdentity
{
    @Column
    private String name;

    @OneToMany(mappedBy = "user")
    private Set<PQ> pqs = new HashSet<PQ>();

    ...
}

组（也是一个 SecurityIdentity 子类，其简单 ID 为 PK，引用 PQ 来表示可以与该 PQ 交互的一组用户）：

@Entity
@Table(name = "Groups")
@DiscriminatorValue(value = "group")
public class Group extends SecurityIdentity
{
    @OneToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "pq_id", referencedColumnName = "id")
    private PQ pq;

    ...
}

招标：

@Entity
@Table(name = "Tenderings")
public class Tendering implements Serializable
{
    @Id
    @Column(name = "pq_id", insertable = false, updatable = false)
    private Integer pqId;

    @Column(name = "external_code")
    private String externalCode;

    @OneToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "pq_id", referencedColumnName = "id")
    private PQ pq;

    ...
}

不要对共享 ID 的组和用户感到困惑，只需将它们视为简单的 ID 即可。招标只是一个单独的文档对象（一对一）。

如您所见，PQ 实体上有三个关系，如果未设置提取类型，则会预先加载（JPA 默认值）。因此，为了防止这种情况，我将所有关系标记为.@XToOne@XToOneFetchType.LAZY

现在使用时

em.find(PQ.class, someExistingId);

我得到休眠输出：

23:53:55,815 INFO  [stdout] Hibernate: select pq0_.id as id291_0_, pq0_.description as descript2_291_0_, pq0_.name as name291_0_, pq0_.submission_date as submission4_291_0_, pq0_.user_id as user5_291_0_ from PQs pq0_ where pq0_.id=?
23:53:55,818 INFO  [stdout] Hibernate: select user0_.id as id280_0_, user0_1_.identity_type_id as identity2_280_0_, user0_.is_enabled as is1_297_0_, user0_.name as name297_0_, user0_.password as password297_0_, user0_.person_id as person5_297_0_ from Users user0_ inner join SecurityIdentities user0_1_ on user0_.id=user0_1_.id where user0_.person_id=?
23:53:55,821 INFO  [stdout] Hibernate: select group0_.id as id280_0_, group0_1_.identity_type_id as identity2_280_0_, group0_.pq_id as pq2_281_0_ from Groups group0_ inner join SecurityIdentities group0_1_ on group0_.id=group0_1_.id where group0_.pq_id=?
23:53:55,823 INFO  [stdout] Hibernate: select tendering0_.pq_id as pq1_296_0_, tendering0_.binary_file as binary2_296_0_, tendering0_.customer_id as customer6_296_0_, tendering0_.description as descript3_296_0_, tendering0_.external_code as external4_296_0_, tendering0_.title as title296_0_ from Tenderings tendering0_ where tendering0_.pq_id=?

另外三个SELECT源于@XToOne关系（如网络上许多地方所描述的那样）。我主要关注的来源是这样的：

使一对一关系变得懒惰

如前所述，不应获取关系：@ManyToOneUser user

@ManyToOne(fetch=FetchType.LAZY)应该工作得很好。

...这里从PQ到用户的关系，但它是从语句中看到的...select user0_.id as id280_0_, ...

对于其他两个和 ，两个反向映射，外键引用 PQs 表的 PK （ID），从而在 PQ 实体中产生相同的映射。Group groupTendering tendering@OneToOne

请注意，所有三种关系都不是可选的：PQ 始终具有所有者（用户），并且 PQ 始终由招标和组实体引用。我只是还没有在上面的JPA中建模...

因此，在添加 PQ 实体的三个关系时：optional = false

@Entity
@Table(name = "PQs")
public class PQ implements Serializable
{
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column
    private Integer id;

    @Column
    private String name;

    @ManyToOne(fetch = FetchType.LAZY, optional = false)
    @JoinColumn(name = "user_id", referencedColumnName = "person_id")
    private User user;

    @OneToOne(mappedBy = "pq", fetch = FetchType.LAZY, optional = false)
    private Group group;

    @OneToOne(mappedBy = "pq", fetch = FetchType.LAZY, optional = false)
    private Tendering tendering;

    ...
}

...我得到以下休眠输出：

00:47:34,397 INFO  [stdout] Hibernate: select pq0_.id as id361_0_, pq0_.description as descript2_361_0_, pq0_.name as name361_0_, pq0_.submission_date as submission4_361_0_, pq0_.user_id as user5_361_0_ from PQs pq0_ where pq0_.id=?
00:47:34,410 INFO  [stdout] Hibernate: select user0_.id as id350_0_, user0_1_.identity_type_id as identity2_350_0_, user0_.is_enabled as is1_367_0_, user0_.name as name367_0_, user0_.password as password367_0_, user0_.person_id as person5_367_0_ from Users user0_ inner join SecurityIdentities user0_1_ on user0_.id=user0_1_.id where user0_.person_id=?
00:47:34,413 INFO  [stdout] Hibernate: select group0_.id as id350_0_, group0_1_.identity_type_id as identity2_350_0_, group0_.pq_id as pq2_351_0_ from Groups group0_ inner join SecurityIdentities group0_1_ on group0_.id=group0_1_.id where group0_.pq_id=?

请注意，我只在PQ实体上玩过，因为这是我在中使用的那个。（如果这还不够，请启发我。optional = falseem.find(...)

我现在的问题是双重的：

1. 为什么对实体的渴望（鉴于据说这是懒惰地工作，请参阅使一对一关系懒惰）？@ManyToOneUser
2. 为什么只提取与实体的关系？是因为实体将 PQ 的 PK 列引用为 PK 本身（in ），而实体没有（与 PQ 的 PK 的常规关系）？OneToOneTenderingTendering@IdTenderingGroup

怎么了？如何使这些非可选关系变得懒惰？（没有代码检测或其他黑客，只有普通的注释...）

我知道LAZY的事情只是JPA提供者对延迟加载或不做一些事情的提示，但是在这种情况下，似乎有其他事情是错误的（因为它的一部分正在工作）。

PS：我使用的是Hibernate 4.0 BETA，JBoss 7.0.0.Final附带的版本以及仅JPA注释（以上都是JPA 1.0兼容的）。