How to start synchronization from program

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Appr3nt1c3 - 23 Aug 2008 04:44 GMT

Hi to everyone.

I have 2 sql servers 2000 running merge replication. Sometimes the
publisher has an error saying that it cannot connect to the
subscriber, and there is a red "x" on the replication agent. What I do
is right click it and click Start Synchronization.

My question is: Is there a way for me to run the "Start
Synchronization" process from within a .Net program.

Thanks in advance.

Ray,

P.S.
I'm new to replication, so please bear with me :)

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Paul Ibison - 23 Aug 2008 17:51 GMT

Presumably you're running the merge agent in continuous mode. The easiest
solution is to run it on a schedule (every 1 min?) instead. I can advise you
on how to initiate the agent programatically if necessary but it might not
be needed.
Rgds,
Paul Ibison (www.replicationanswers.com)

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Appr3nt1c3 - 24 Aug 2008 07:47 GMT

> Presumably you're running the merge agent in continuous mode. The easiest
> solution is to run it on a schedule (every 1 min?) instead. I can advise you
> on how to initiate the agent programatically if necessary but it might not
> be needed.
> Rgds,
> Paul Ibison (www.replicationanswers.com)

Hi Paul,

Thanks for your reply.

Yes it is running in continuous mode. I'll set in on schedule and see
what happens.

I want to know how to run it from a program, cause i have an sms modem
installed. When the users inform me that they could not see the
updates from their side, i can send an sms to this modem and have a
program process the message and run the agent from within the code.
This is in case i'm out of the office, as i'm frequenty am.

Thanks,

Ray

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Paul Ibison - 24 Aug 2008 09:23 GMT

OK - for the programmatic synchronization, if you need it, please take a
look here: http://support.microsoft.com/kb/319649
HTH,
Paul Ibison (www.replicationanswers.com)

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Appr3nt1c3 - 24 Aug 2008 11:50 GMT

> OK - for the programmatic synchronization, if you need it, please take a
> look here:http://support.microsoft.com/kb/319649
> HTH,
> Paul Ibison (www.replicationanswers.com)

Thanks. I'll look at the code.

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