Tom Lane wrote:
>
> Joseph Shraibman <[EMAIL PROTECTED]> writes:
> > Stephan Szabo wrote:
> >> Do you have a value that is not null that is very common?
> >> It's estimating that there will be 10113 rows that match
> >> nomsession='xxx' which makes a seq scan a much less bad plan.
> >>
> > Err, why? There is an index, isn't there? Shouldn't the index allow
> > postgres to quickly find the %2 of rows that would match?
>
> Define "quickly".
>
> > sitefr=# explain select nomsession from session where nomsession='xxx';
> > NOTICE: QUERY PLAN:
> >
> > Seq Scan on session (cost=0.00..16275.95 rows=10113 width=12)
>
> We have here an estimate that 10113 rows will be matched (out of the
> 510069 in the table). The table contains something on the order of
> 16000 pages (guesstimate from the seqscan cost estimate). The planner
> is assuming that the 10113 rows are randomly scattered in the table,
> and therefore that the executor will have to fetch the majority of the
> pages in the table. Under these circumstances a seqscan is cheaper
> than an indexscan, because it works with the Unix kernel's preference
> for sequential reads (to say nothing of the disk drive's ;-)), instead
> of fighting that optimization. Random fetches are more than twice as
> expensive as sequential fetches.
>
> Of course, if the 10113-match estimate is wildly off (as it was in this
> case), then the wrong plan may be chosen. But it IS NOT CORRECT to
> suppose that indexscans always beat seqscans. The planner's job would
> be a lot easier if that were true.
>
> regards, tom lane
Can't postgres do the index lookup first and find out there are only a
few tuples that might match?
--
Joseph Shraibman
[EMAIL PROTECTED]
Increase signal to noise ratio. http://www.targabot.com
