Yo man,
all you've gotto do is to write your own UserType for Scala Enums (please
post it as Open Source for all to benefit)
Cheers,
Viktor
On Fri, Sep 19, 2008 at 11:03 PM, Charles F. Munat <[EMAIL PROTECTED]> wrote:
>
> I'm still trying to get a Gender enumeration to work with JPA. The
> problem now seems to be on the JPA end (I'm using the JPADemo code that
> Derek provided).
>
> I get the following error:
> javax.persistence.PersistenceException: org.hibernate.MappingException:
> Could not determine type for: scala.Enumeration$Value, for columns:
> [org.hibernate.mapping.Column(gender)]
>
> Here is the relevant code:
>
>
> @serializable
> object Gender extends Enumeration {
> type Gender = Value
> val Unknown = Value(0, "Unknown")
> val Male = Value(1, "Male")
> val Female = Value(2, "Female")
> }
>
> import Gender._
>
> @Entity
> @Table{val name = "USERS"}
> class User {
> @Id
> @GeneratedValue{val strategy = GenerationType.AUTO}
> var id : Long = _
>
> @Column{val name="NAME_LAST"}
> var nameLast : String = ""
>
> @Column{val name="NAME_FIRST"}
> var nameFirst : String = ""
>
> var username : String = ""
>
> @Column{val name="EMAIL_ADDRESS"}
> var emailAddress : String = ""
>
> @Enumerated(EnumType.ORDINAL)
> var gender : Gender = _
>
> ...
> }
>
> Also, where is the best place to put the Gender enumeration so that it
> is visible both in the persistence project and in the webapp?
>
> Thanks for any and all help.
>
> Chas.
>
> >
>
--
Viktor Klang
Senior Systems Analyst
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