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Yes, only when the message has a DLQ header it processes the message and performs the action performed.
Here's a sample dlq script that i used long back.
Quote "
INPUTQ('DEADQ') INPUTQM('FRED')
REASON(*) ACTION(FWD) FWDQ('EMPTYQ') +
FWDQM('FRED') RETRY(3)
"Unquote
Remember to add the carriage return explicitly after the rule.
Cheers
Kumar
-------Original Message-------
Date: Monday, June 23, 2003 02:44:45 PM
Subject: Re: Why runmqdlq is not doing its job?
How did you put the messages on the dlq? I believe the handler ignores them
if they don't have a dead letter header.
Bill
William M. Pigg
Bill Pigg/San Jose/IBM
408.256.5656 t/l 276.5656
cc:
Subject: Why runmqdlq is not doing its job?
I am on MQ 5.3 CSD03 on W2K. I have a qmgr QM1 which
uses QM1.DEAD as its dead letter q. Both QM1.DEAD and
Q1 are defined as local queues on QM1 (they both are
put/get enabled, maxmsglenghth = qmgr max length,
qdepth=40000, Shared). Currently QM1.DEAD has 3
messages on it.
I have a rule table (mydlq.rul) that just has these 2
lines in it just for testing:
inputqm(QM1) inputq(QM1.DEAD)
ACTION(FWD) FWDQ(Q1) FWDQM(QM1) HEADER(YES)
When I run the runmqdlq on the command prompt as:
Runmqdlq c:\mydlq.rul
It starts running but QM1.DEAD still has 3 messages,
as if the runmqdlq is ignoring the action(fwd) in
the rule table. Why is it not working?
Thanks,
Ruzi
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