休眠：如何使用父對象的自動生成的ID作為外鍵來級聯嵌入式對象

Question

題：

在Hibernate（JPA 2.0）中，創建父對象時如何創建具有@OneToOne關系的嵌入式對象？

期望：

當我創建用戶時，我希望它會自動創建適當的屬性，並將用戶的自動生成的id作為屬性的object_id 。

現實：

當我嘗試創建用戶時，出現以下異常：

javax.persistence.PersistenceException: org.hibernate.PropertyValueException: not-null property references a null or transient value: com.api.user.PropertyEntity.objectId

數據庫表：

我有兩個MySQL表： users和properties

users
+----+----------+----------+
| id | username | password |
+----+----------+----------+
| 1  | foo      | bar      |
+----+----------+----------+
| 2  | goo      | baz      |
+----+----------+----------+
| 3  | woo      | hah      |
+----+----------+----------+

properties
+----+-----------+-------------+---------+-----------+
| id | object_id | object_type | propkey | propvalue |
+----+-----------+-------------+---------+-----------+
| 1  | 1         | user        | sso     | true      |
+----+-----------+-------------+---------+-----------+
| 2  | 1         | user        | ssoid   | foobar    |
+----+-----------+-------------+---------+-----------+
| 3  | 2         | user        | sso     | false     |
+----+-----------+-------------+---------+-----------+
| 4  | 2         | user        | ssoid   | null      |
+----+-----------+-------------+---------+-----------+
| 5  | 3         | user        | sso     | false     |
+----+-----------+-------------+---------+-----------+
| 6  | 3         | user        | ssoid   | null      |
+----+-----------+-------------+---------+-----------+

Java實體：

一個用戶可以具有多個屬性，但是每個屬性都與該用戶具有@OneToOne關系，並且該關系是單向的（即，用戶創建其屬性，而不是相反）。

Java中的User實體如下所示：

@Entity
@Table(name = "users")
public class UserEntity {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id", nullable = false)
    private Integer id;

    @Column(name = "username", nullable = false)
    private String userName;

    @Column(name = "password", nullable = false)
    private String password;

    @OneToOne
    @Cascade(value = { org.hibernate.annotations.CascadeType.SAVE_UPDATE, org.hibernate.annotations.CascadeType.DELETE })
    @JoinColumnsOrFormulas(value={
            @JoinColumnOrFormula(column=
                   @JoinColumn(name = "id", referencedColumnName = "object_id", insertable=false, updatable=false, nullable = false)
            ),
            @JoinColumnOrFormula(formula=
                    @JoinFormula(referencedColumnName="object_type", value="'user'")
            ),
            @JoinColumnOrFormula(formula=
                    @JoinFormula(referencedColumnName="propkey", value="'sso'")
            )
    })
    private PropertyEntity isSso;

    @OneToOne
    @Cascade(value = { org.hibernate.annotations.CascadeType.SAVE_UPDATE, org.hibernate.annotations.CascadeType.DELETE })
    @JoinColumnsOrFormulas(value={
            @JoinColumnOrFormula(column=
                    @JoinColumn(name = "id", referencedColumnName = "object_id", insertable=false, updatable=false, nullable = false)
            ),
            @JoinColumnOrFormula(formula=
                    @JoinFormula(referencedColumnName="object_type", value="'user'")
            ),
            @JoinColumnOrFormula(formula=
                    @JoinFormula(referencedColumnName="propkey", value="'ssoid'")
            )
    })
    private PropertyEntity ssoId;

    // Getters & Setters
}

Property實體如下所示：

@Entity
@Table(name = "properties")
public class PropertyEntity {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id", nullable = false)
    private Integer id;

    @Column(name = "object_id", nullable = false)
    private Integer objectId;

    @Column(name = "object_type", nullable = false)
    private String objectType;

    @Column(name = "propkey", nullable = false)
    private String name;

    @Column(name = "propvalue", nullable = false)
    private String value;

    // Getters & Setters
}

Answer 1

像大多數與Hibernate注釋有關的問題一樣，我通過完全消除它們並以老式的方式來解決此問題（在我的事務中進行單獨的查詢以創建/更新用戶，然后創建/更新每個用戶屬性記錄）。

如果任何人都有可以使用的Hibernate解決方案，我很樂意接受該答案。