On Wed, Mar 15, 2006 at 09:06:06PM -0800, Andrew Ng wrote:
> replicated to the slave, but the sequence's last value is still not
> updated at the slave. As such, subsequent inserts would fail with
I don't think so. The snapshots are applied in a transaction. So
the setval() actually happens in the same transaction: you wouldn't
see the data at all. The danger, of course, is that you lose data on
failover. That's a known risk with failover, which is why there's
such an emphasis on doing controlled switchover, if possible. Don't
wait until the machine's dead: if there's evidence of trouble on your
origin, switch over sooner.
A
--
Andrew Sullivan | [EMAIL PROTECTED]
In the future this spectacle of the middle classes shocking the avant-
garde will probably become the textbook definition of Postmodernism.
--Brad Holland
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