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how to send XML message with userid to start workflow proces |
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mqwf1 |
Posted: Fri Jan 18, 2002 3:17 am Post subject: |
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Novice
Joined: 13 Jan 2002 Posts: 22
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I have find way to send message to EXE.XML to start WorkFlow process,
but I don't know how to set tag to tell WorkFlow Runtime who
is the starter of the process instance.
where can find this XML DTD ?
Pls tell me what should I do.
Thanks in advance.
AnDong
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yoscop |
Posted: Mon Jan 28, 2002 6:25 am Post subject: |
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Novice
Joined: 17 Dec 2001 Posts: 13
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hi,
We can set in the openOptions as MQC.MQOO_SET_IDENTITY_CONTEXT and then explicitly set MQMessage.userid="Anyuser valid MQWFUser".
The following code snippet might help.
int openOptions = MQC.MQOO_INPUT_AS_Q_DEF |
MQC.MQOO_OUTPUT | MQC.MQOO_SET_IDENTITY_CONTEXT ;
// Now specify the queue that we wish to open,
// and the open options...
com.ibm.mq.MQQueue putQueue = qMgr.accessQueue(xmlInputQ, // EXEXMLINPUTQ
openOptions);
// null, // default q manager
// null, // no dynamic q name
// null); // no alternate user id
// prepare the data to put
MQMessage msg2put = new MQMessage();
msg2put.userId = wfAdmin;
msg2put.replyToQueueName = _ReceiveFromQName;
msg2put.replyToQueueManagerName = wfQMgr;
msg2put.writeChars(putXmlMsg);
// specify the message options...
MQPutMessageOptions pmo = new MQPutMessageOptions(); // accept the // defaults,
pmo.options = MQC.MQPMO_SET_IDENTITY_CONTEXT; // new code
// same as MQPMO_DEFAULT
// put the message on the queue
putQueue.put(msg2put,
pmo);
Regards,
yoscop |
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