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Home/ Questions/Q 6681341
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T04:34:45+00:00 2026-05-26T04:34:45+00:00

The factory will be based on Default messaging provider . The reason for doing

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The factory will be based on Default messaging provider.

The reason for doing so is I have a mock MDB in place of an external system, normally connected via Websphere MQ. The mock and the tested application are installed on the same server, so I need to make sure that in the cycle

request -> mock listener -> mock reply -> response

request and reply are two separate transactions.

I can’t see any option in the console to remove the factory from transaction manager’s supervision.

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  1. Editorial Team
    Editorial Team
    2026-05-26T04:34:46+00:00Added an answer on May 26, 2026 at 4:34 am

    You are correct in your observation that the JMS Connection Factory cannot be configured to not enlist in the current transaction. I am a little unsure of what you are trying to achive here, but I’m going to make the following assumption:

    1. Client sends message
    2. Listener picks up message and sends reply
    3. Client picks up the reply

    In this scenario the client needs to send and commit the message before the listener will get it to send the reply. The listener can receive and reply in one transaction, but the client needs two. I am furthermore assuming that the client is an EJB.

    One solution to this would be use bean managed transactions. In this case you can manually begin a transaction and commit it around 1 and again around 3. This will cause the message to be sent.

    Another solution would be to do the send in an EJB method with RequiresNew and the receive in another EJB method with RequiresNew. Then the client calls the send EJB method followed by the receive EJB method and each method has its own transaction.

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