Question

我试图通过Hibernate插入并选择mysql数据库中的数据，并且Insert对我来说工作正常，但是选择某种方式不要映射正确的表名并且不会返回结果。

获取并插入代码：

    SessionFactory sessFact = HibernateUtil.getSessionFactory();
    Session session = sessFact.getCurrentSession();
    session.beginTransaction();

    session.save(obj);
    session.getTransaction().commit();

    try {
        Session mysession = HibernateUtil.getSessionFactory().getCurrentSession();
        mysession.beginTransaction();
        weatherDataObject resultObjectHib = (weatherDataObject) mysession.get(weatherDataObject.class, 26);
        mysession.getTransaction().commit();
    } catch (Exception e) {
        e.printStackTrace();
    }
    sessFact.close();



<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
        "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
        "http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">
<hibernate-mapping package="some.pack">
    <class name="weatherCoordinates" table="coordinates">
        <id name="dataBaseId" column="coordinates_id">
            <generator class="native" />
        </id>
        <property name="lat" type="string" column="coordinates_lat" />
        <property name="lon" type="string" column="coordinates_lon" />




    </class>
</hibernate-mapping>




<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
        "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
        "http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">
<hibernate-mapping package="some.pack">
    <class name="weatherDataObject" table="data_object">
        <id name="id" column="data_object_id">
            <generator class="native" />
        </id>
        <property name="name" type="string" column="data_object_name" />
        <many-to-one name="coord" class="task.main.DataObjects.weatherCoordinates"
            column="coordinates_id" unique="true" not-null="true" cascade="all" />

    </class>
</hibernate-mapping>

当我看到sql执行字符串时，它是：

Hibernate: insert into coordinates (coordinates_lat, coordinates_lon) values (?, ?)
Hibernate: insert into data_object (data_object_name, coordinates_id) values (?, ?)
Hibernate: select weatherdat0_.data_object_id as data_obj1_1_0_, weatherdat0_.data_object_name as data_obj2_1_0_, weatherdat0_.coordinates_id as coordina3_1_0_ from data_object weatherdat0_ where weatherdat0_.data_object_id=?

问题是weatherdat0我的表格在某种程度上是错误的，因为我将其映射到文件data_object中的方式不知道如何以及为什么更改任何人可以提供帮助？

Answer 1

查询是在正确的表上生成的，因为select查询在表from data_object上运行

weatherdat0_只是from语句中提到的表的别名：

from data_object weatherdat0_

所以它只选择正确的表名。

现在如果查询没有返回任何结果意味着没有匹配id的记录，所以我建议你直接在数据库上运行查询，看它是否返回任何记录。

休眠选择错误的表名

1 个答案: