Question

我有两个hibernate映射文件。我已经尝试了几种使用此documentation来绑定它们的方法。你能告诉我最简单的方法吗？

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
        "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <class name="model.Player" table="PLAYER">
        <id name="playerId" type="int" >
            <column name="ID" precision="5" scale="0"/>
            <generator class="sequence">
                <param name="sequence">PLAYER_SEQ</param>
            </generator>
        </id>
        <!-- ... -->
    </class>
</hibernate-mapping>

第二个：

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
        "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <class name="model.AnnualFee" table="ANNUAL_FEE">
        <id name="id" type="int" >
            <column name="ID" precision="5" scale="0"/>
            <generator class="increment"/>
        </id>
        <property name="playerId" type="int">
            <column name="PLAYER_ID" length="20" not-null="true" />
        </property>
        <!-- ... -->
    </class>
</hibernate-mapping>

我希望它们与playerId相关联，并且能够进一步执行outer join。

编辑：

我想这样做：

<id name="playerId" type="int" >
            <column name="ID" precision="5" scale="0"/>
            <generator class="sequence">
                <param name="sequence">PLAYER_SEQ</param>
            </generator>
        </id>

        <set name="playerId" table="ANNUAL_FEE">
            <key column="PLAYER_ID"/>
            <one-to-many class="model.AnnualFee" />
        </set>

但是，然后想法会突出显示getPlayerId()：

“为字段＆get; getPlayerId＆＃39;”
配置了多个属性

Answer 1

<many-to-one name="player" insert="false" update="false" class="model.Player" fetch="select">
    <column name="ID" not-null="true" />
</many-to-one>

添加到AnnualFee不只是int playerId而是Player

public class AnnualFee {
    private int id;
    private Player player;
//...

与hibernate的简单关联映射

1 个答案: