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Home -> Community -> Mailing Lists -> Oracle-L -> Re: a explain plan question
At 07:43 AM 1/30/2007, amonte wrote:
>Hi
>
>I have this query
SNIP
>and this plan
>
>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)
>
>--------------------------------------------------------------------------------------------
>| Id | Operation | Name | Rows | Bytes |
>Cost | Pstart| Pstop |
>--------------------------------------------------------------------------------------------
>| 0 | SELECT STATEMENT | | 934 | 49502
>| 1022K| | |
>| 1 | SORT GROUP BY | | 934 | 49502
>| 1022K| | |
>| 2 | NESTED
>LOOPS | | 182M| 9213M| 814K| | |
>|* 3 | TABLE ACCESS FULL | TUD_FEDIA | 30 | 480
>| 3 | | |
>| 4 | PARTITION RANGE
>ITERATOR| | | | | KEY | KEY |
>|* 5 | TABLE ACCESS FULL | TUH_NVPAGINA | 5992K| 211M|
>27159 | KEY | KEY |
>--------------------------------------------------------------------------------------------
SNIP
>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?
Unless I'm completely mistaken, the first step is to access the driving table of the nested loop join, i.e. step 3. Besides, if it didn't it wouldn't have the information for the filter predicate for step 5
An interesting side-observation (for me at least) was that the M stands for MB as in 1,048,576=1024*1024, not Million as in 1,000,000. I wouldn't intuitively count rows in multiples of 1024.
Regards
Wolfgang Breitling
Centrex Consulting Corporation
www.centrexcc.com
-- http://www.freelists.org/webpage/oracle-lReceived on Tue Jan 30 2007 - 09:22:28 CST
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