BizTalk Server General - many BizTalk 2006 applications sending to the same sql server

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Author many BizTalk 2006 applications sending to the same sql server
jemkem

2006-06-03, 7:20 pm

Hi all!
I'm using BizTalk 2006.

I have several orchestrations that all end up sending data to the same
database. I thought I'd create a BT 2006 application corresponding to
the sql server destination, make a host for that app, and make send
ports to send data into the sql server. Then just reference the set of
send ports built in the sql server application. Good application
segmentation. Then I can add a second sql server application for a
duplicate environment, and bind orchs to one or the other sql server
application's send ports. Also, be able to reset the host instance on
the sql server app, without disturbing anything else.

There are 2 problems (at least, right?). One is, my new host doesn't
appear in the pick list for the send handler on my send ports. The
other is that I can't reference ports between applications.

Am I heading down the wrong path?

Thanks,
Jay

jemkem

2006-06-23, 1:18 pm

Nobody has anything to offer on this one?

jemkem wrote:
> Hi all!
> I'm using BizTalk 2006.
>
> I have several orchestrations that all end up sending data to the same
> database. I thought I'd create a BT 2006 application corresponding to
> the sql server destination, make a host for that app, and make send
> ports to send data into the sql server. Then just reference the set of
> send ports built in the sql server application. Good application
> segmentation. Then I can add a second sql server application for a
> duplicate environment, and bind orchs to one or the other sql server
> application's send ports. Also, be able to reset the host instance on
> the sql server app, without disturbing anything else.
>
> There are 2 problems (at least, right?). One is, my new host doesn't
> appear in the pick list for the send handler on my send ports. The
> other is that I can't reference ports between applications.
>
> Am I heading down the wrong path?
>
> Thanks,
> Jay


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