Re: [Springframework-developer] access to (LocalSessionFactoryBean).createDatabaseSchema()

[<jue...@we...>]

You need to fetch the LocalSessionFactoryBean itself here, rather than =
the object it exposes:

  LocalSessionFactoryBean lsfb =3D (LocalSessionFactoryBean) =
ctx.getBean("&mySessionFactory");

This is rarely necessary, but works with any FactoryBean: Simply prepend =
a "&" to the bean name.

Juergen


-----Original Message-----
From: spr...@li...
[mailto:spr...@li...]On Behalf
Of Jozsa Kristof
Sent: Wednesday, June 02, 2004 10:28 AM
To: spr...@li...
Subject: [Springframework-developer] access to
(LocalSessionFactoryBean).createDatabaseSchema()


Hi,

I'd like to create my database schema with my configured=20
LocalSessionFactoryBean's convenient method createDatabaseSchema(). I=20
configured my LSFB the usual way:

<bean id=3D"sessionFactory"=20
class=3D"org.springframework.orm.hibernate.LocalSessionFactoryBean">
   <property name=3D"dataSource"><ref local=3D"oracleDS"/></property>
   ...
</bean>

..but if I pull that bean 'sessionFactory' from the spring container, it =

will be a type of net.sf.hibernate.SessionFactoryImpl, the default=20
implementation of the hibernate SessionFactory interface. How could I=20
invoke the method mentioned to create the database schema then? (I can't =

really see either how the resulting bean could be a total different type =

than the configured one, but that has minor priority compared to my=20
original question).

As a side note, I never really understood why Hibernate doesn't let=20
access the used Configuration from the SessionFactory instance, which=20
could be use to feed SchemaExport and create the schema..

tia,
dyn
--=20
.Digital.Yearning.for.Networked.Assassination.and.Xenocide.


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