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Hi, am getting the following error while doing an schemaexport. Here are my part of my code :

/**
* Get a set of parameters for this Application
* @hibernate.set
* inverse = "true"
* order-by = "application_id"
* @hibernate.collection-one-to-many
* class = "com.gloptv.Parameter"
*/
public Set getParameters()
{
return this.parameters;
}

And part of my Application.hmb.xml generated file
<set
name="parameters"
lazy="false"
inverse="true"
cascade="none"
sort="unsorted"
order-by="application_id"
>

<one-to-many
class="com.gloptv.Parameter"
/>
</set>


Can anyone please help me solve my problem
thanks in advance.

[b]Hibernate version:3.0[/b]

[b]Mapping documents:[/b]

[b]Code between sessionFactory.openSession() and session.close():[/b]

[b]Full stack trace of any exception that occurs:
[schemaexport] (util.XMLHelper 59 ) Error parsing XML: M:\tomcat\webapps\smsc\WEB-INF\classes\com\gloptv\Application.h
bm.xml(66) The content of element type "set" must match "(meta*,subselect?,cache?,synchronize*,key,(element|one-to-many|many-to-many|composi
te-element|many-to-any),loader?,sql-insert?,sql-update?,sql-delete?,sql-delete-all?,filter*)".
[/b]

[b]Name and version of the database you are using:postgresql 8.0[/b]

[b]The generated SQL (show_sql=true):[/b]

[b]Debug level Hibernate log excerpt:[/b]

You may want to read the error - it is an XML validation error - you are missing the <key> element in the <set> element.

thanks. it works now