Question

已经在这方面工作了几天，还没有看到一个与我尝试做的相符的例子，或者我错过了它，因为我有点新鲜Hibernate和JPA。我试图将一些hibernate代码转换为JPA，并且似乎无法使特定连接正确。

这是我的表结构：

表AppUser


id（PK）


表安全问题


id（PK）


表AppUserSecurityQuestion



AppUserId（PK，FK to AppUser.id）



SecurityQuestionId（PK，FK to SecurityQuestion.id）

以下是我尝试过的域名（只是相关的财产声明）：

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;
    ...
}

这是在使用hibernate配置，但我又试图将其转换为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>

我基本上只是试着查看hbm并将每个部分转换为JPA，但是当我尝试通过应用程序访问数据时，我显然遗漏了一些内容：

org.hibernate.exception.SQLGrammarException：无法提取ResultSet ...由以下原因引起： com.microsoft.sqlserver.jdbc.SQLServerException：无效的对象名称＆＃39; AppUser_AppUserSecurityQuestion＆＃39;

这也是HQL：

选择securityan0_.AppUser_id作为AppUser_1_0_0_， securityan0_.securityAnswers_appUser为security2_3_0_， securityan0_.securityAnswers_securityQuestion as security3_3_0_， appusersec1_.appUser为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和 securityan0_.securityAnswers_securityQuestion = appusersec1_.securityQuestion securityan0_.AppUser_id =？

我显然已经错误地宣布了我的联接，但我不确定此时还有什么可以尝试。有谁看到我做错了什么？

Answer 1

想出来......至少它似乎按照我需要的方式工作。以下是我需要做出的更改：

<强> 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;
    ...
}

<强> AppUserSecurityQuestionId.java

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

其他一切都保持不变。

JPA注释复合主键，它也是一对多的外键

1 个答案: