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


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Andrew Sullivan  | ajs at crankycanuck.ca
In the future this spectacle of the middle classes shocking the avant-
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