Hello, I don't know how to declare an intermediate subclass in Hibernate mapping file adding few properties, can someone help ?

I'm using a table-per-hierarchy strategy and so discriminator-value to map between different types of subclasses.

Here is the case (Hibernate version: 2.1.4) :

I'd like to have a base class Datasource declared as :



Then few concrete classes directly extending it :



And then to add key & query fields :


And next extending WithKeyAndQueryDatasource to concrete classes :



All I found is to forget the usage of intermediate WithKeyAndQueryDatasource class and have the following mapping, it works, but doesn't look the most beautiful mapping file ever, see red section repeated three times.

Mapping documents:
<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-2.0.dtd" >

<hibernate-mapping package="com.mycompany.app">
<class name="Datasource" table="Tbl_Datasource">
<composite-id class="DatasourcePK" name="primaryKey">
<key-property
column="Id_Datasource"
name="idDatasource"
type="string"
/>
<key-many-to-one
class="Exhibit"
column="Id_Exhibit"
name="exhibit"
/>
</composite-id>
<discriminator
column="Cd_DatasourceType"
type="string"
/>
<subclass name="ManualInputDatasource" discriminator-value="MAN"/>
<subclass name="FixedDatasource" discriminator-value="FIX"/>
<subclass name="CSVDatasource" discriminator-value="CSV">
<property
column="Txt_key"
length="255"
name="key"
not-null="false"
type="string"
/>
<property
column="Cd_Query"
length="255"
name="query"
not-null="false"
type="string"
/>
</subclass>
<subclass name="G2Datasource" discriminator-value="GE2">
<property
column="Txt_key"
length="255"
name="key"
not-null="false"
type="string"
/>
<property
column="Cd_Query"
length="255"
name="query"
not-null="false"
type="string"
/>
</subclass>
<subclass name="V2Datasource" discriminator-value="EF2">
<property
column="Txt_key"
length="255"
name="key"
not-null="false"
type="string"
/>
<property
column="Cd_Query"
length="255"
name="query"
not-null="false"
type="string"
/>
</subclass>
</class>
</hibernate-mapping>

Whereas I'd like to be able to declare many discriminator values, so to have something like :


How can I do this ? Any Idea ?
Thanx, r0main

Hi!

why can't you continue to declare subclasses within subclasses? Similar to your proposed mapping? Is there a problem with abstract classes?

Hi, thanks for the reply...
The fact is, in your suggested declaration :

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-2.0.dtd" >

<hibernate-mapping package="com.mycompany.app">
<class name="Datasource" table="Tbl_Datasource">
<composite-id class="DatasourcePK" name="primaryKey">
<key-property column="Id_Datasource" name="idDatasource" type="string"/>
<key-many-to-one class="Exhibit" column="Id_Exhibit" name="exhibit"/>
</composite-id>

<discriminator column="Cd_DatasourceType" type="string" />

<subclass name="ManualInputDatasource" discriminator-value="MAN"/>
<subclass name="FixedDatasource" discriminator-value="FIX"/>
<subclass name="WithKeyAndQueryDatasource" discriminator-value="WKQ">
<property column="Txt_key" length="255" name="key" not-null="false" type="string" />
<property column="Cd_Query" length="255" name="query" not-null="false" type="string" />

<subclass name="CSVDatasource" discriminator-value="CSV"/>
<subclass name="G2Datasource" discriminator-value="GE2"/>
<subclass name="V2Datasource" discriminator-value="EF2"/>
</subclass>
</class>
</hibernate-mapping>

My db is so I cannot add another discriminator column and am forced to use the "Cd_DatasourceType" one (definition is in BLUE).

The real problem is we may need to have more than one possible value for a given discriminator related to a class.

So in your declaration, the line :


may become :


If only a multiple-values discriminator switch was permitted...

Romain