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hibernate mapping of subclass and individual fields

开发者 https://www.devze.com 2023-02-03 20:52 出处:网络
I have this class hierachy: class A { private String a1; private Class B a2; // getter & setter } class B

I have this class hierachy:

class A
{
  private String a1;
  private Class B a2;
// getter & setter
}

class B
{
  private String b1;
  private String b2;
  // getter & setter
}
class C extends class A
{
  private String c1;
  // getter and setter for c1;
}

And I need to map C to a single table in a databas开发者_StackOverflowe using hibernate. I try this:

<class name="com.C" 
    table="myTable" catalog="myCatalog"
    polymorphism="implicit">
    <property name="a1" type="string">
            <column name="column1" length="40" not-null="true" />
    </property>
    <property name="a2" type="????">
            ?? B should map to column2 and column3
    </property>
    <property name="c1" type="string">
            <column name="column4" length="40" not-null="true" />
    </property>
</class>

How to map field b of class B?

Thanks,


Hi found it using component tag:

<class name="com.C" 
    table="myTable" catalog="myCatalog"
    polymorphism="implicit">
    <property name="a1" type="string">
            <column name="column1" length="40" not-null="true" />
    </property>
    <component name="a2" class="com.B">
        <property name="b1" type="java.lang.Integer">
            <column name="column2" />
        </property>
        <property name="b2" type="java.lang.Integer">
            <column name="column3" />
        </property>        
    </component> 
    <property name="c1" type="string">
            <column name="column4" length="40" not-null="true" />
    </property>
</class>
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