RE: a explain plan question
- From: "Juan Miranda" <j.miranda@xxxxxxxxxxxx>
- To: <Bernard.Polarski@xxxxxxxxxxxxxx>, <oracle-l@xxxxxxxxxxxxx>
- Date: Wed, 31 Jan 2007 10:39:49 +0100
> 4 2 PARTITION RANGE (ITERATOR)
> 5 4 TABLE ACCESS (FULL) OF 'TUH_NVPAGINA' (Cost=27159
Means that Oracle are using some partitions, and then are doing a full scan
of these partitions.
Alex, I think you must create an LOCAL partitioned index on
TUH_NVPAGINA.FE_DIA.
If you have more columns that FE_DIA in your primary key, Oracle may not use
it.
You need some like this:
2 1 PARTITION RANGE (ITERATOR)
3 2 INDEX (RANGE SCAN) OF 'TUH_NVPAGINA_IDX1' (INDEX)
greetings
_____
De: oracle-l-bounce@xxxxxxxxxxxxx [mailto:oracle-l-bounce@xxxxxxxxxxxxx] En
nombre de Polarski, Bernard
Enviado el: miércoles, 31 de enero de 2007 10:24
Para: ax.mount@xxxxxxxxx; oracle-l@xxxxxxxxxxxxx
Asunto: RE: a explain plan question
Still I don?t understand why we have a PARTITION RAND followed by a full
table scan. Why not a direct full table scan, what is the advantage of this
construct
> 4 2 PARTITION RANGE (ITERATOR)
> 5 4 TABLE ACCESS (FULL) OF 'TUH_NVPAGINA' (Cost=27159
?
On 1/30/07, LS Cheng < exriscer@xxxxxxxxx> wrote:
TUD_FEDIA is accessed first then from that it eliminates partitions
(partition start/stop KEY), the problem seems nested loop, how many rows is
TU_FEDIA returning?
?
I don?t see the keyword(stop key) in the plan. I read this plan and its only
speak of a partition range access that leads to a full table scan.
My only explanation is that the CBO is underlining a failed partition
pruning.
Bernard Polarski
_____
From: amonte [mailto:ax.mount@xxxxxxxxx]
Sent: woensdag 31 januari 2007 10:09
To: oracle-l@xxxxxxxxxxxxx
Subject: Re: a explain plan question
You are correct, the NL is not good, I changed to hash join and the query
runs in 50 minutes.
Thanks
Alex
On 1/30/07, LS Cheng < exriscer@xxxxxxxxx> wrote:
doesnt look very good plan
TUD_FEDIA is accessed first then from that it eliminates partitions
(partition start/stop KEY), the problem seems nested loop, how many rows is
TU_FEDIA returning?
On 1/30/07, Remigiusz Soko?owski < rems@xxxxxxxx <mailto:rems@xxxxxxxx> >
wrote:
>
> Execution Plan
> ----------------------------------------------------------
> 0 SELECT STATEMENT Optimizer=CHOOSE (Cost=1022392 Card=934
> Bytes=49502)
> 1 0 SORT (GROUP BY) (Cost=1022392 Card=934 Bytes=49502)
> 2 1 NESTED LOOPS (Cost=814767 Card=182275095 Bytes=9660580035)
> 3 2 TABLE ACCESS (FULL) OF 'TUD_FEDIA' (Cost=3 Card=30
> Bytes=480)
> 4 2 PARTITION RANGE (ITERATOR)
> 5 4 TABLE ACCESS (FULL) OF 'TUH_NVPAGINA' (Cost=27159
> Card=5992606 Bytes=221726422)
>
> I was wondering how to read this plan, the order of steps. From old
> set autotrace trace exp it seems to me that step 5 is the first step?
>
> 5 4 TABLE ACCESS (FULL) OF 'TUH_NVPAGINA' (Cost=27159
> Card=5992606 Bytes=221726422)
>
AFAIK the first most nested line is the first line (in this example the
one indicated by You)
Regards
Remigiusz
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Remigiusz Sokolowski <rems@xxxxxxxx <mailto:rems@xxxxxxxx> >
WP/PTI/DIP/ZAB (+04858) 52 15 770
MySQL v. 4.x
Oracle v. 10.x
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