Question

我的代码是这样的：我想在两个表格中选择数据

    // ArrayList<DetailType> list = null;
    SessionFactory sf = null;
    Session ses = null;
    try {

        Configuration cfg = new Configuration().configure();
        ServiceRegistry sr = new ServiceRegistryBuilder().applySettings(
                cfg.getProperties()).buildServiceRegistry();
        sf = cfg.buildSessionFactory(sr);
        ses = sf.openSession();
        // 执行查询语句

        String hql = "select new DetailVoMain(d.d_id,d.d_name,m.m_name)"
                + "from MainType m, DetailType d "
                + "where d.d_main_id = m.m_id";
        Query query = ses.createQuery(hql);

        List<?> list = query.list();
        for (int i = 0; i < list.size(); i++) {
            DetailVoMain detailType = (DetailVoMain) list.get(i);
            System.out.println("dId-------->" + detailType.getdId());
            System.out.println("dNam-------->" + detailType.getdName());
            System.out.println("mNam-------->" + detailType.getmName());
        }
    } catch (HibernateException e) {
        System.out.println("hibernate exception");
        e.printStackTrace();

    } finally {
        ses.close();
        sf.close();
    }

和myeclipese给我错误信息：

hibernate exception

org.hibernate.QueryException：无法解析属性：d_id：com.xunfang.em_mallServer.bean.DetailType [从com.xunfang.em_mallServer中选择新的DetailVoMain（d.d_id，d.d_name，m.m_name）。 bean.MainType m，com.xunfang.em_mallServer.bean.DetailType d其中d.d_main_id = m.m_id] 在org.hibernate.persister.entity.AbstractPropertyMapping.propertyException（AbstractPropertyMapping.java:83）在org.hibernate.persister.entity.AbstractPropertyMapping.toType（AbstractPropertyMapping.java:77）在org.hibernate.persister.entity.AbstractEntityPersister.toType（AbstractEntityPersister.java:1968）在org.hibernate.hql.internal.ast.tree.FromElementType.getPropertyType（FromElementType.java:313）在org.hibernate.hql.internal.ast.tree.FromElement.getPropertyType（FromElement.java:490）在org.hibernate.hql.internal.ast.tree.DotNode.getDataType（DotNode.java:616）在org.hibernate.hql.internal.ast.tree.DotNode.prepareLhs（DotNode.java:267）在org.hibernate.hql.internal.ast.tree.DotNode.resolve（DotNode.java:214）在org.hibernate.hql.internal.ast.tree.FromReferenceNode.resolve（FromReferenceNode.java:119）在org.hibernate.hql.internal.ast.tree.FromReferenceNode.resolve（FromReferenceNode.java:115）在org.hibernate.hql.internal.ast.tree.DotNode.resolveSelectExpression（DotNode.java:678）在org.hibernate.hql.internal.ast.HqlSqlWalker.resolveSelectExpression（HqlSqlWalker.java:893）在org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectExpr（HqlSqlBaseWalker.java:2208）在org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.constructor（HqlSqlBaseWalker.java:2502）在org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectExpr（HqlSqlBaseWalker.java:2275）在org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectExprList（HqlSqlBaseWalker.java:2145）在org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectClause（HqlSqlBaseWalker.java:1451）在org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.query（HqlSqlBaseWalker.java:571）在org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectStatement（HqlSqlBaseWalker.java:299）在org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.statement（HqlSqlBaseWalker.java:247）在org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze（QueryTranslatorImpl.java:249）在org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile（QueryTranslatorImpl.java:184）在org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile（QueryTranslatorImpl.java:137）在org.hibernate.engine.query.spi.HQLQueryPlan。（HQLQueryPlan.java:105）在org.hibernate.engine.query.spi.HQLQueryPlan。（HQLQueryPlan.java:80）在org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan（QueryPlanCache.java:168）在org.hibernate.internal.AbstractSessionImpl.getHQLQueryPlan（AbstractSessionImpl.java:221）在org.hibernate.internal.AbstractSessionImpl.createQuery（AbstractSessionImpl.java:199）在org.hibernate.internal.SessionImpl.createQuery（SessionImpl.java:1778）在com.xunfang.em_mallServer.util.Test.main（Test.java:39）

请帮助我，我整夜测试！

我的MainType实体是这样的：

public class MainType {
private String mId; // 编号
private String mName; // 类别名称

public String getmId() {
    return mId;
}

public void setmId(String mId) {
    this.mId = mId;
}

public String getmName() {
    return mName;
}

和这样的映射：

<hibernate-mapping package="com.xunfang.em_mallServer.bean">

<class name="MainType" table="t_main_type"> <!-- 设置表名和字段对应数据 -->
    <id name="mId" column="m_id" type="string">

    </id>
    <property name="mName" column="m_name" type="string"></property>

</class>

DetailType实体，如下所示：

public class DetailType {
/**
 * 小类编号
 */
private String dId;
/**
 * 小类名称
 */
private String dName;
/**
 * 归属大类编号
 */
private String dMainId;

public String getdId() {
    return dId;
}

public void setdId(String dId) {
    this.dId = dId;
}

public String getdName() {
    return dName;
}

public void setdName(String dName) {
    this.dName = dName;
}

public String getdMainId() {
    return dMainId;
}

public void setdMainId(String dMainId) {
    this.dMainId = dMainId;
}

和映射：

<hibernate-mapping package="com.xunfang.em_mallServer.bean">

<class name="DetailType" table="t_detail_type"> <!-- 设置表名和字段对应数据 -->
    <id name="dId" column="d_id" type="string">

    </id>
    <property name="dName" column="d_name" type="string"></property>
    <property name="dMainId" column="d_main_id" type="string"></property>

</class>

谢谢你的帮助！

Answer 1

你应该写HQL正确，使用映射属性，例如dId（而不是d_id）。
您需要在实体之间添加关系（例如OneToMany，ManyToOne）之后，您可以加入hql中的其他实体。

如何使用hibernate hql在两个表中选择数据？

1 个答案: