hibernate - java.sql.SQLException : Parameter index out of range (3 > number of parameters,为2)

我有一个表“AuthorFollow”，其复合键“authorId”和“userId”分别是“AuthorInfo”和“UserInfo”表中的主键。

我正在尝试在数据库中保存“AuthorFollow”的对象(我正在使用mysql)。
但是，我得到了错误:

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 (?, ?)

最佳答案

来自Hibernate的错误消息不清楚，这使得很难查看问题出在哪里。这就是为什么我给了你+1。

您的错误是:在AuthorFollow的映射中不正确。您映射author_id和user_id两次。 Hibernate无法解决这个问题。在插入中，计算参数时失败。每列应仅映射一次(除非一个映射是只读的，但是这种特殊情况与您的问题无关)。

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AuthorInfo或UserInfo实例，则必须加载单独的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)中使用更复杂的解决方案有很多用途或其他良好原因。解决方案2)肯定是比较麻烦的。

关于hibernate - java.sql.SQLException : Parameter index out of range (3 > number of parameters,为2)，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/16122958/