Hibernate：如何使用父对象的自动生成ID作为外键级联嵌入式对象

Question

问题：

在Hibernate（JPA 2.0）中，在创建父对象时，如何创建具有@OneToOne关系的嵌入式对象？

期望值：

当我创建一个用户时，我希望它能自动创建相应的属性，将用户自动生成的id作为属性的对象id。

现实：

当我尝试创建用户时，我得到以下异常：

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

数据库表：

我有两个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实体：

一个用户可以有多个属性，但每个属性都与该用户有@OneToOne关系，并且这种关系是单向的（即用户创建其属性，而不是相反）。

Java中的用户实体如下所示：

@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
}

属性实体如下所示：

@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

与大多数与Hibernate注释相关的问题一样，我解决了这个问题，方法是完全消除它们，并以老式的方式进行（在事务中分别查询创建/更新用户，然后创建/更新每个用户属性记录）。

如果有人有一个有效的Hibernate解决方案，我很乐意接受这个答案。