java - JPA注释复合主键也是一对多外键

Question

我已经为此工作了几天，但还没有看到与我正在尝试做的事情相匹配的示例，或者我错过了它，因为我对 Hibernate 和 JPA 有点陌生。我正在尝试将一些 hibernate 代码转换为 JPA，但似乎无法正确获取特定的连接。

这是我的表结构:

Table AppUser

• id (PK)

Table SecurityQuestion

• id (PK)

Table AppUserSecurityQuestion

• AppUserId (PK, FK to AppUser.id)

• SecurityQuestionId (PK, FK to SecurityQuestion.id)

这是我对我的域所做的尝试(只是相关的属性声明):

BaseEntity.java

@MappedSuperclass
public abstract class BaseEntity implements Serializable {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name="id", updatable=false, nullable=false)
    private Integer id;
    ...
}

AppUser.java

@Entity
@Table(name="AppUser")
public class AppUser extends BaseEntity {
    @OneToMany(fetch=FetchType.EAGER)
    @MapKeyJoinColumn(name="AppUserId")
    private Set<AppUserSecurityQuestion> securityAnswers;
    ...
}

AppUserSecurityQuestion.java

@Entity
@Table(name="AppUserSecurityQuestion")
public class AppUserSecurityQuestion implements java.io.Serializable {
    @EmbeddedId
    private AppUserSecurityQuestionId id;
    ...
}

AppUserSecurityQuestionId.java

@Embeddable
public class AppUserSecurityQuestionId implements java.io.Serializable {
    private AppUser appUser;
    private SecurityQuestion securityQuestion;
    ...
}

这适用于 hibernate 配置，但我再次尝试将其转换为 JPA。以下是 hbm 文件的相关部分:

AppUser.hbm.xml

<hibernate-mapping default-access="field">
    <class catalog="WEBR" name="testapp.domain.AppUser" schema="dbo" table="AppUser">
        <id name="appUserId" type="java.lang.Integer">
            <column name="AppUserId" />
            <generator class="org.hibernate.id.enhanced.SequenceStyleGenerator">
                <param name="sequence_name">AppUserSeq</param>
            </generator>
        </id>
        ...
        <set name="securityAnswers" table="AppUserSecurityQuestion" inverse="true" lazy="false">
            <key column="AppUserId" not-null="true" />
            <one-to-many class="testapp.domain.AppUserSecurityQuestion" />
        </set>


    </class>
</hibernate-mapping>

AppUserSecurityQuestion.hbm.xml

<hibernate-mapping>
    <class name="testapp.domain.AppUserSecurityQuestion" table="AppUserSecurityQuestion" schema="dbo" catalog="TEST">
        <composite-id name="id" class="testapp.domain.AppUserSecurityQuestionId">
            <key-many-to-one name="appUser" class="testapp.domain.AppUser">
                <column name="id"/>
            </key-many-to-one>
            <key-many-to-one name="securityQuestion" class="testapp.domain.SecurityQuestion">
                <column name="SecurityQuestionId" />
            </key-many-to-one>
        </composite-id>
        <property name="answer" type="java.lang.String">
            <column name="Answer" not-null="true" />
        </property>
    </class>
</hibernate-mapping>

我基本上只是想查看 hbm 并将每个部分转换为 JPA，但当我尝试通过应用程序访问数据时出现此异常时，我显然错过了一些东西:

org.hibernate.exception.SQLGrammarException: could not extract ResultSet ... Caused by: com.microsoft.sqlserver.jdbc.SQLServerException: Invalid object name 'AppUser_AppUserSecurityQuestion'

这也是 HQL:

select securityan0_.AppUser_id as AppUser_1_0_0_, securityan0_.securityAnswers_appUser as security2_3_0_, securityan0_.securityAnswers_securityQuestion as security3_3_0_, appusersec1_.appUser as appUser1_2_1_, appusersec1_.securityQuestion as security2_2_1_, appusersec1_.Answer as Answer3_2_1_ from AppUser_AppUserSecurityQuestion securityan0_ inner join AppUserSecurityQuestion appusersec1_ on securityan0_.securityAnswers_appUser=appusersec1_.appUser and securityan0_.securityAnswers_securityQuestion=appusersec1_.securityQuestion where securityan0_.AppUser_id=?

我显然错误地声明了我的连接，但我不确定此时还可以尝试什么。有人看到我做错了什么吗？

Answer 1

想通了...至少它看起来按照我需要的方式工作。以下是我需要进行的更改:

AppUser.java

@Entity
@Table(name="AppUser")
public class AppUser extends BaseEntity {
    @OneToMany(mappedBy="id.appUser", fetch=FetchType.EAGER)
    @MapKeyJoinColumn(name="id")
    private Set<AppUserSecurityQuestion> securityAnswers;
    ...
}

AppUserSecurityQuestionId.java

@Embeddable
public class AppUserSecurityQuestionId implements java.io.Serializable {
    @ManyToOne
    @JoinColumn(name="AppUserId")
    private AppUser appUser;
    @ManyToOne
    @JoinColumn(name="SecurityQuestionId")
    private SecurityQuestion securityQuestion;
    ...
}

其他一切保持原样。