> Chip Wilson wrote:
> > 2) The sequence diagrams for entity beans with BMP show the underlying
> > persistent store, represented by "database" on the diagram, registering
> > itself with the same transaction service controlled by the
> > javax.jts.UserTransaction.  How can this be?  My persistent store may be
> a
> > legacy system that knows nothing about EJB, how could it know to
> register
> > itself with the server's transaction mechanism?
>
        Rickard wrote:
> If you want the bean to be transactional, even though you're using BMP,
> your underlying store must be able to hook itself into the transaction
> manager. It can do this by following the instructions in the JTA spec
> (in short: the JTA impl. must register an interface object in JNDI which
> resources can use to receive transactional notifications).
>
[Chip Wilson]
If this is true, then there are several corollaries:
*       Your persistent store must be coupled to your particular EJB server,
and must be blessed by your server vendor, even if you are using BMP.  This
is implied by the fact that JTA/JTS is not required by the EJB spec.  So,
when you choose an EJB server, you must choose a persistent store from the
list of supported ones for that server.  BMP does not allow you to use just
any persistence mechanism.
*       If your persistent store is not variable, then you must choose a
server which supports it.
*       You cannot use BMP to integrate EJB with a legacy persistent store
or application.  The persistent store or application vendor must provide a
version that is coupled with your particular EJB server.  This would seem to
contradict the statement in the spec that "In many cases, the underlying
data source may be an existing application rather than a database."
*       You cannot use BMP with an off-the-shelf O/R mapping framework,
since these typically implement their own transaction mechanism, unaware of
JTS/JTA/EJB.

I was under the impression that one motivation for BMP was to accommodate
all of the existing, "legacy" applications out there, but this does not
appear to be true.  Comments?

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