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List:       ms-dcom
Subject:    Re: What transaction level do MTS/DTC transactions run under?
From:       Scott Prugh <sprugh () TELUTION ! COM>
Date:       1999-04-30 17:17:22
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>MTS/COM+ Declarative Transactions run at ISOLATIONLEVEL_SERIALIZABLE.

>SQL Server picks this up at enlistment time. Trying to change it using
>ADO/OLEDB won't work (SQLS ignores the ado connection's isolation level).
>You CAN tweak using stuff like SELECT NOLOCK and friends on a statement by
>statement basis.

>DB
http://www.develop.com/dbox


If this is true, then is it fair to say that all tables involved in a Tx are
locked?  In my previous example, that would mean that both T1 and T2 will be
locked for the duration of the transaction.  Since this type of lock would
incur a key range lock in Sql Server, you take a huge hit on concurrency for
inserts.  Basically, all the inserts single-thread.

Additionally, where did you find this info?  Is there no other way to tweak
this behavior?


Regards,
        Scott






> -----Original Message-----
> From: Scott Prugh [mailto:sprugh@TELUTION.COM]
> Sent: Friday, April 30, 1999 7:38 AM
> To: DCOM@DISCUSS.MICROSOFT.COM
> Subject: What transaction level do MTS/DTC transactions run under?
>
>
> I'm trying to figure which transaction level MTS components
> run under when
> updating Sql Server tables with ADO.
>
> Component A(Requires New):
>
> A::UpdateTables()
> {
>
> // Update Table T1
> // Update Table T2
>
> SetComplete();
> }
>
>
> Now, when A::UpdateTables is called, MTS enlists DTC for a new
> transaction(TX1) and both T1 and T2 are updated.under that
> transaction.
>
> What types of locking occur here?  Are both T1, and T2 locked
> during TX1?
> If another thread(or activity) tries to read T1(or T2) before TX1 is
> commited/rolled back then what results are seen?  in standard
> client server
> apps, I can set the isolation
> level(SERIALIZABLE,READ_COMMITED,READ_UNCOMMITTED).
> Unfortunately, I have
> not been able to find any info of what occurs under a DTC
> transaction in
> MTS?
>
> Does anyone have any ideas on this?
>
>
> Regards,
>         Scott

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