I'm having problems wrapping my head around this to come with a suitable solution. I inherited this database that has a couple of tables that need to be mapped together. I discovered that composite keys must have the same number of columns (through a stack track seen below). Short question: Is this possible to match on a single column of a composite key? Long question: Could somebody point me to some code that could give me an example of this?

TIA,
Tino

PS: If you need anything else, please ask

Hibernate version: 3.2.5

Mapping documents:
Fullname.hbm.xml
<hibernate-mapping>
<class name="Fullname" table="fullname" schema="someSchema">
<composite-id name="id" class="FullnameId">
<key-property name="id" type="java.lang.Long">
<column name="ID" precision="5" scale="0" />
</key-property>
<key-property name="fromDate" type="java.util.Date">
<column name="FROM_DATE" length="7" />
</key-property>
</composite-id>
<property name="lastname" type="java.lang.String">
<column name="LASTNAME" length="160" not-null="true" />
</property>
<property name="firstname" type="java.lang.String">
<column name="FIRSTNAME" length="120" not-null="true" />
</property>
... fields deleted ......
<set name="Mhscode">
<key column="id"/>
<one-to-many class="Mhscode"/>
</set>
</class>
</hibernate-mapping>
mhscode.hbm.xml

<hibernate-mapping>
<class name="gov.loc.legis.hibernate.Mhscode" table="MHSCODE" schema="LISBSS">
<composite-id name="id" class="gov.loc.legis.hibernate.MhscodeId">
<key-property name="id" type="java.lang.Long">
<column name="ID" precision="5" scale="0" />
</key-property>
<key-property name="code" type="java.lang.String">
<column name="CODE" length="40" />
</key-property>
<key-property name="fromDate" type="java.util.Date">
<column name="FROM_DATE" length="7" />
</key-property>
</composite-id>
..... stuff deleted .......
<!-- unsure if I need another mapping back to fullname -->
</class>
</hibernate-mapping>

Stack trace:

%%%% Error Creating SessionFactory %%%%
org.hibernate.MappingException: Foreign key (FK68CE0405EF493335:MHSCODE [id])) must have same number of columns as the referenced primary key (fullname [ID,FROM_DATE])
at org.hibernate.mapping.ForeignKey.alignColumns(ForeignKey.java:90)
at org.hibernate.mapping.ForeignKey.alignColumns(ForeignKey.java:73)
at org.hibernate.cfg.Configuration.secondPassCompileForeignKeys(Configuration.java:1263)
at org.hibernate.cfg.Configuration.secondPassCompile(Configuration.java:1170)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1286)
at HibernateSessionFactory.<clinit>(HibernateSessionFactory.java:31)
at BaseHibernateDAO.getSession(BaseHibernateDAO.java:13)
at FullnameDAO.findByProperty(FullnameDAO.java:93)
at FullnameDAO.findByFirstname(FullnameDAO.java:107)
at TestSponsorFetchDriver.main(TestSponsorFetchDriver.java:22)


database: Oracle 9.2.0.8 (yes, I know it is old!)

Ciao,

Short response: Yes it is!

Long Response:

Go to this link: http://forum.hibernate.org/viewtopic.php?t=991318&highlight= and look how I suggested to map the composite-id in a separate class and how after that you can create a criteria over a single field from the composite-id.

Hope it helps.