Hibernate:映射3个表

时间:2009-05-12 11:57:58

标签: hibernate mapping

我正在尝试使用Hibernate映射一些现有的表。

这很简单:我们的类别有多种语言的名称。

DDL如下:

create table language (
   id                integer not null auto_increment,
   code              varchar(2) not null,

   unique (code),

   primary key(id)
);

create table category (
   id               integer not null auto_increment,
   parent_id        integer default null,
   ordr             integer not null default 99,

   primary key (id)
);

create table category_description (
   category_id      integer not null,
   language_id      integer not null,

   title            varchar(255) not null,

   constraint foreign key (category_id) references category(id),
   constraint foreign key (country_language_id) references country_language(id),

   primary key (category_id, country_language_id)
);

现在我想要一个带有语言的地图,因为它的关键和描述(表category_description)就像它的值一样,如下所示:

private Map<Language, CategoryDescription> descriptions = new HashMap<Language, CategoryDescription>();

任何人都可以向我提供一些指示吗?我已经尝试了第311/312页中给出的'Java Persistence with Hibernate'中的示例,它类似于我的问题,但我只是没有得到它:(

1 个答案:

答案 0 :(得分:2)

(您的DDL不一致,您创建表“语言”但引用表“country_language” - 我将假设后者)

以下是样本的Hibernate映射:

<?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 default-lazy="false">

    <class name="Language" table="country_language">
        <id name="id" type="int">
            <column name="ID" />
            <generator class="native" />
        </id>
        <property name="code" type="string" length="2" unique="true" />
    </class>

    <class name="Category" table="category">
        <id name="id" type="int">
            <column name="ID" />
            <generator class="native" />
        </id>
        <map name="descriptions" table="category_description">
            <key column="category_id" />
            <map-key-many-to-many column="language_id" class="Language" />
            <composite-element class="CategoryDescription">
                <property name="title" type="string" length="255" />
            </composite-element>
        </map>
    </class>

</hibernate-mapping>

但是,您根本不需要CategoryDescription类(因为它只包含String):

private Map<Language, String> descriptions;

<map name="descriptions" table="category_description">
    <key column="category_id" />
    <map-key-many-to-many column="language_id" class="Language" />
    <element type="string" length="255" column="title" />
</map>

也可以。

请注意,在这两种情况下,Language类都需要覆盖hashCode()equals()才能成功查询生成的地图:

/* eclipse generated */
@Override
public int hashCode() {
    final int prime = 31;
    int result = 1;
    result = prime * result + ((id == null) ? 0 : id.hashCode());
    return result;
}

/* eclipse generated */
@Override
public boolean equals(Object obj) {
    if (this == obj)
        return true;
    if (obj == null)
        return false;
    if (getClass() != obj.getClass())
        return false;
    Language other = (Language) obj;
    if (id == null) {
        if (other.id != null)
            return false;
    } else if (!id.equals(other.id))
        return false;
    return true;
}
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