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Home -> Community -> Mailing Lists -> Oracle-L -> Re: Mysterious FILTER operation ;)
The statement in the performance tuning guide is wrong, and the suggestion it makes has only ever been a rough guideline.
This particular type of FILTER was present in (at least) 7.2, (this specific FILTER appeared only in 9i, it doesn't appear in 8i).
For a more traditional example of a plan where the statement is wrong, consider a simple nested loop:
nested loop
table t1 full
table t2 by rowid
index unique t2_pk (unique)
The first action is the start of the scan of table T1, it is NOT the rightmost action - which is the index access into t2_pk.
Regards
Jonathan Lewis
http://www.jlcomp.demon.co.uk/faq/ind_faq.html The Co-operative Oracle Users' FAQ
http://www.jlcomp.demon.co.uk/seminar.html Public Appearances - schedule updated Dec 23rd 2004
This is very interesting observation that I would like to discuss.
According to the "Performance Tuning Guide", for example, the
execution order begins with the line that is the furthest indented to
the right
http://download-west.oracle.com/docs/cd/B14117_01/server.101/b10752/optimops.htm#73843
But this experiment (v_p1>v_p2) ) proves that real execution order can be different. If we have LIO equal to 0, than INDEX RANGE SCAN was not performed and FILTER was performed before RANGE SCAN, this is also proved by sql_trace execution plan:
Rows Row Source Operation
------- --------------------------------------------------- 1 SORT AGGREGATE (cr=0 pr=0 pw=0 time=57 us) 0 FILTER (cr=0 pr=0 pw=0 time=8 us) 0 TABLE ACCESS BY INDEX ROWID OBJ#(282311) (cr=0 pr=0 pw=0 time=0 us) 0 INDEX RANGE SCAN OBJ#(282328) (cr=0 pr=0 pw=0 time=0us)(object id 282328)
As we can see from time column to FILTER step consumed 8 us but INDEX RANGE SCAN consumed 0 us (was not executed).
Can somebody share some thoughts about real execution path of the statement?
-- http://www.freelists.org/webpage/oracle-lReceived on Tue Jan 04 2005 - 15:44:51 CST
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