java.sql.SQLException:参数索引超出范围(3>参数个数,即2)

Question

我有一个表"AuthorFollow",它具有'authorId'和'userId'的复合键,它们分别是"AuthorInfo"和"UserInfo"表中的主键.

我试图在数据库中保存"AuthorFollow"的对象(我正在使用mysql).但是,我收到了错误:

org.hibernate.exception.GenericJDBCException: could not insert: [com.pojo.hibernate.AuthorFollow]
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Caused by: java.sql.SQLException: Parameter index out of range (3 > number of parameters, which is 2).
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我正在尝试保存对象的代码是:

Transaction transaction = hibernateTemplate.getSessionFactory().getCurrentSession().beginTransaction();
    try {
        AuthorFollow authorFollow = new AuthorFollow();
        authorFollow.setAuthorId(authorInfo.getAuthorId());
        authorFollow.setUserId(userInfo.getUserId());
        authorFollow.setAuthorInfoByAuthorId(authorInfo);
        authorFollow.setUserInfoByUserId(userInfo);

        authorInfo.getAuthorFollowsByAuthorId().add(authorFollow);
        userInfo.getAuthorFollowsByUserId().add(authorFollow);

        updateObject(authorInfo);
        updateObject(userInfo);

        transaction.commit();
        return true;
    } catch (Exception e) {
        e.printStackTrace();
        transaction.rollback();
        return false;
    }

映射文件(这些映射由IntellijIdea自动生成):AuthorFollow.hbm.xml:

<hibernate-mapping>
    <class name="com.pojo.hibernate.AuthorFollow" table="author_follow" catalog="book">
    <composite-id mapped="true" class="com.pojo.hibernate.AuthorFollowPK">
        <key-property name="userId" column="user_id"/>
        <key-property name="authorId" column="author_id"/>
    </composite-id>
    <many-to-one name="authorInfoByAuthorId" class="com.pojo.hibernate.AuthorInfo">
        <column name="author_id" not-null="true"/>
    </many-to-one>
    <many-to-one name="userInfoByUserId" class="com.pojo.hibernate.UserInfo">
        <column name="user_id" not-null="true"/>
    </many-to-one>
    </class>
</hibernate-mapping>

AuthorInfo.hbm.xml(仅显示AuthorFollow的映射):

<set name="authorFollowsByAuthorId" inverse="true">
        <key>
            <column name="author_id" not-null="true"/>
        </key>
        <one-to-many not-found="ignore" class="com.pojo.hibernate.AuthorFollow"/>
    </set>

UserInfo.hbm.xml(仅显示AuthorFollow的映射):

<set name="authorFollowsByUserId" inverse="true">
        <key>
            <column name="user_id" not-null="true"/>
        </key>
        <one-to-many not-found="ignore" class="com.pojo.hibernate.AuthorFollow"/>
    </set>

更新:

Hibernate: select userinfo0_.user_id as user1_21_, userinfo0_.first_name as first2_21_, userinfo0_.last_name as last3_21_, userinfo0_.user_gender as user4_21_, userinfo0_.user_img as user5_21_, userinfo0_.user_birthdate as user6_21_, userinfo0_.user_occupation as user7_21_, userinfo0_.user_qualification as user8_21_, userinfo0_.user_postal_code as user9_21_, userinfo0_.user_address as user10_21_, userinfo0_.user_city as user11_21_, userinfo0_.user_contact as user12_21_, userinfo0_.user_balance as user13_21_, userinfo0_.user_website as user14_21_, userinfo0_.email_verified as email15_21_, userinfo0_.email_id as email16_21_ from book.user_info userinfo0_ where userinfo0_.email_id=?
Hibernate: select authorinfo0_.author_id as author1_3_0_, authorinfo0_.author_name as author2_3_0_, authorinfo0_.author_pen_name as author3_3_0_, authorinfo0_.author_gender as author4_3_0_, authorinfo0_.author_description as author5_3_0_, authorinfo0_.author_blog_link as author6_3_0_, authorinfo0_.author_img as author7_3_0_, authorinfo0_.author_lives as author8_3_0_, authorinfo0_.author_born as author9_3_0_, authorinfo0_.author_died as author10_3_0_, authorinfo0_.author_notable_works as author11_3_0_ from book.author_info authorinfo0_ where authorinfo0_.author_id=?
Hibernate: select userinfo0_.user_id as user1_21_, userinfo0_.first_name as first2_21_, userinfo0_.last_name as last3_21_, userinfo0_.user_gender as user4_21_, userinfo0_.user_img as user5_21_, userinfo0_.user_birthdate as user6_21_, userinfo0_.user_occupation as user7_21_, userinfo0_.user_qualification as user8_21_, userinfo0_.user_postal_code as user9_21_, userinfo0_.user_address as user10_21_, userinfo0_.user_city as user11_21_, userinfo0_.user_contact as user12_21_, userinfo0_.user_balance as user13_21_, userinfo0_.user_website as user14_21_, userinfo0_.email_verified as email15_21_, userinfo0_.email_id as email16_21_ from book.user_info userinfo0_ where userinfo0_.user_id=?
Hibernate: select authorfoll0_.author_id as author2_3_1_, authorfoll0_.user_id as user1_1_, authorfoll0_.author_id as author2_1_, authorfoll0_.user_id as user1_1_0_, authorfoll0_.author_id as author2_1_0_ from book.author_follow authorfoll0_ where authorfoll0_.author_id=?
Hibernate: select authorfoll0_.user_id as user1_21_1_, authorfoll0_.user_id as user1_1_, authorfoll0_.author_id as author2_1_, authorfoll0_.user_id as user1_1_0_, authorfoll0_.author_id as author2_1_0_ from book.author_follow authorfoll0_ where authorfoll0_.user_id=?
Hibernate: insert into book.author_follow (author_id, user_id) values (?, ?)

Answer 1

来自Hibernate的错误消息不明确,这使得很难看出问题所在.这就是我给你+1的原因.

您的错误是:在映射中AuthorFollow不正确.您映射author_id和user_id的两倍.Hibernate无法处理这个问题.在插入中,它在计算参数时失败.每列只应映射一次(除非一个映射是只读的,但特殊情况对您的问题不感兴趣).

有两种解决方案:

1)简单的解决方案:使用id代替AuthorInfo和UserInfo对象:

<hibernate-mapping>
  <class name="com.pojo.hibernate.AuthorFollow" table="author_follow" catalog="book">
    <composite-id mapped="true" class="com.pojo.hibernate.AuthorFollowPK">
      <key-property name="userId" column="user_id"/>
      <key-property name="authorId" column="author_id"/>
    </composite-id>
  </class>
</hibernate-mapping>

没有<many-to-one>属性了.如果您明确需要一个AuthorInfo或UserInfo实例,则必须加载一个单独的Session.load()语句.

2)复杂的解决方案:使用<key-many-to-one>:

<hibernate-mapping>
  <class name="com.pojo.hibernate.AuthorFollow" table="author_follow" catalog="book">
    <composite-id mapped="true" class="com.pojo.hibernate.AuthorFollowPK">
      <key-many-to-one name="authorInfoByAuthorId" class="com.pojo.hibernate.AuthorInfo" column="author_id"/>
      <key-many-to-one name="userInfoByUserId" class="com.pojo.hibernate.UserInfo" column="user_id"/>
    </composite-id>
  </class>
</hibernate-mapping>

我建议使用简单的解决方案1),除非2)中有更多复杂解决方案的大量使用或其他充分理由.解决方案2)明确地说更麻烦.