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Editorial Team
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Asked: 2026-05-27T01:26:06+00:00

I’d like to set the v$session.program Oracle property in order to have information available

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I’d like to set the v$session.program Oracle property in order to have information available in the session table. I’m using JPA with a jndi XA datasource created with an oracle-xa-ds.xml deployed in the deploy folder of Jboss, and therefore I haven’t access to the constructor of the Connection.
I have access to the Connection object, in JPA 2 using unwrap, in JPA 1 by casting JPA to Hibernate classes, but there are no properties setter (only Client Info properties that are the way to proceed starting JDBC 4.0).

So my question is, using JPA (with Hibernate) using Jboss 4.2 :

  • Is it possible to configure the v$session.program in the persistence.xml ?
  • Is it possible to configure the v$session.program in the oracle-ds.xml ?
  • Is their any other approach to the solution ?

Thank you for any valuable comments and answers !

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1 Answer

  1. Editorial Team

    I had the same Problem today, after much fiddeling and reading documentation finally I had the Eureka moment:

    Add following parameter:

    <xa-datasource-property name="connectionProperties">v$session.program=YourUniqueName</xa-datasource-property>

    Thats all.

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