From: | Markus Schiltknecht <markus(at)bluegap(dot)ch> |
---|---|
To: | Simon Riggs <simon(at)2ndquadrant(dot)com> |
Cc: | NikhilS <nikkhils(at)gmail(dot)com>, pgsql-hackers(at)postgresql(dot)org |
Subject: | Re: Auto Partitioning |
Date: | 2007-04-04 14:31:54 |
Message-ID: | 4613B6DA.9010908@bluegap.ch |
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Lists: | pgsql-hackers pgsql-patches |
Hi,
Simon Riggs wrote:
> I agree with much of your post, though this particular point caught my
> eye. If you'll forgive me for jumping on an isolated point in your post:
No problem.
> Multi-table indexes sound like a good solution until you consider how
> big they would be. The reason we "need" a multi-table index is because
> we are using partitioning, which we wouldn't be doing unless the data
> was fairly large. So the index is going to be (Num partitions *
> fairly-large) in size, which means its absolutely enormous. Adding and
> dropping partitions also becomes a management nightmare, so overall
> multi-table indexes look unusable to me. Multi-table indexes also remove
> the possibility of loading data quickly, then building an index on the
> data, then adding the table as a partition - both the COPY and the
> CREATE INDEX would be slower with a pre-existing multi-table index.
I agree. (And thanks to TOAST, we never have very wide tables with
relatively few rows, right? I mean, something like pictures stored in
bytea columns or some such.)
> My hope is to have a mechanism to partition indexes or recognise that
> they are partitioned, so that a set of provably-distinct unique indexes
> can provide the exact same functionlity as a single large unique index,
> just without the management nightmare.
Uhm... I don't quite get what you mean by "provably-distinct unique
indexes".
As long as the first columns of an index are equal to all columns of the
partitioning columns, there is no problem. You could easily reduce to
simple per-table indexes and using the partitioning rule set to decide
which index to query.
But how to create an (unique) index which is completely different from
the partitioning key?
Regards
Markus
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