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MQSeries.net Forum Index » Multiphase Commit » xa_open() concept - who calls this routine ?

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sebastia
PostPosted: Sun Mar 31, 2013 3:34 am    Post subject: xa_open() concept - who calls this routine ? Reply with quote

Grand Master

Joined: 07 Oct 2004
Posts: 1003

Hi. I have a XA concept shadow ...

In XA spec document

>>> http://pubs.opengroup.org/onlinepubs/009680699/toc.pdf

... I read ...

Quote:
In each thread of control, the TM must call xa_open() for each RM directly accessible by that thread before calling any other xa_ routine.


So, I understand TM is calling xa_open() for all RM's

But when queue manager starts, in AMQERR01.LOG we get a message

Quote:
AMQ7605: The XA resource manager bwbd56 has returned an unexpected return code -5, when called for xa_open.
EXPLANATION: WebSphere MQ received an unexpected return code when calling XA resource manager bwbd56 at its xa_open entry point.


So, the sequence of events has been

1) queue manager starts
2) somehow TM is aware of this fact
3) TM calls RM "bwbd56" (Oracle in fact) with mq's XA_Open_String from qm.ini

I have 2 questions, at least :

1) how does TM know mq has started ?
2) how does TM receive XAOpenString from QM.INI ?

Sebastian.
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bruce2359
PostPosted: Sun Mar 31, 2013 6:22 pm    Post subject: Re: xa_open() concept - who calls this routine ? Reply with quote

Poobah

Joined: 05 Jan 2008
Posts: 9394
Location: US: west coast, almost. Otherwise, enroute.

sebastia wrote:
1) how does TM know mq has started ?

At qmgr restart, the qmgr (RM) registers with the TM. Search the document you referenced for 'register.'
sebastia wrote:
2) how does TM receive XAOpenString from QM.INI ?
IMS, it flows as part of the qmgr (RM) registration process.
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