a plan explain time column
HelloI use Oracle version 10.2.0.3.0. I have 2 tables with 10 million documents each. The DOF is as follows.
create the bigtable table (varchar2 (20) col1, col2 varchar2 (20))
create table bigtablechild (varchar2 (20) col1, col2 varchar (20))
bigtablechild.Col1 is a foreign key to bigtable.col1. Below is the request and explain the plan. During several executions, the query runs for about 20 seconds before returning the results. Could someone please explain what represents the time column? It does not at the same time it took return of results.
SQL> set autotrace on
SQL>
SQL> select b.col2
2 from bigtable a, bigtablechild b
3 where a.col1 = b.col1
4 and a.col1 = 'ABC6554';
COL2
--------------------
XYZ6554
XYZ6554
XYZ6554
XYZ6554
XYZ6554
Execution Plan
----------------------------------------------------------
Plan hash value: 4210396901
------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 5 | 150 | 21538 (4)| 00:04:19 |
|* 1 | HASH JOIN | | 5 | 150 | 21538 (4)| 00:04:19 |
|* 2 | TABLE ACCESS FULL| BIGTABLE | 1 | 10 | 13124 (4)| 00:02:38 |
|* 3 | TABLE ACCESS FULL| BIGTABLECHILD | 5 | 100 | 8413 (5)| 00:01:41 |
------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - access("A"."COL1"="B"."COL1")
2 - filter("A"."COL1"='ABC6554')
3 - filter("B"."COL1"='ABC6554')
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
93672 consistent gets
91845 physical reads
0 redo size
463 bytes sent via SQL*Net to client
396 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
5 rows processedHello
>
Thanks, but how did you determine the 12 Member States based on my exit from the plan to explain?
04:19 = 259 seconds = 259 000 milliseconds
259 000 milliseconds / 21538 monobloc bed ~ = 12 milliseconds per read monobloc
Note, however, that A time column, on the other hand, is extremely useful, but it is available only if content plan (RowSource) statistics have been populated.What is the A-time column, you are referring to? Could you please give an example?
ALTER session set statistics_level = all;
Select * from table (dbms_xplan.display_cursor (
Best regards
Nikolai
Tags: Database
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(hope this execution plan is reasonably legible)--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | Id | Operation | Name | Starts | E-Rows | A-Rows | A-Time | Buffers | Reads | OMem | 1Mem | Used-Mem | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | 0 | SELECT STATEMENT | | 1 | | 155 |00:06:36.28 | 4957K| 34952 | | | | | 1 | SORT ORDER BY | | 1 | 1 | 155 |00:06:36.28 | 4957K| 34952 | 55296 | 55296 |49152 (0)| | 2 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.04 | 4957K| 34952 | | | | | 3 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.04 | 4957K| 34952 | | | | | 4 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.04 | 4957K| 34952 | | | | | 5 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.04 | 4956K| 34952 | | | | | 6 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.04 | 4956K| 34952 | | | | | 7 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.03 | 4956K| 34952 | | | | | 8 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.03 | 4956K| 34952 | | | | | 9 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.03 | 4956K| 34952 | | | | | 10 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.03 | 4955K| 34952 | | | | | 11 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.03 | 4955K| 34952 | | | | | 12 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.03 | 4955K| 34952 | | | | | 13 | NESTED LOOPS | | 1 | 1 | 155 |00:06:30.03 | 4954K| 34952 | | | | |* 14 | HASH JOIN | | 1 | 1 | 155 |00:06:30.03 | 4954K| 34952 | 872K| 872K| 927K (0)| | 15 | VIEW | | 1 | 15 | 8 |00:06:28.63 | 1305K| 34883 | | | | |* 16 | FILTER | | 1 | | 8 |00:06:28.63 | 1305K| 34883 | | | | | 17 | HASH GROUP BY | | 1 | 15 | 8 |00:06:28.63 | 1305K| 34883 | 760K| 760K| 1077K (0)| | 18 | VIEW | | 1 | 15 | 341 |00:00:50.44 | 1305K| 34883 | | | | | 19 | UNION-ALL | | 1 | | 341 |00:00:50.44 | 1305K| 34883 | | | | | 20 | VIEW | V_POSNR_2011000 | 1 | 7 | 303 |00:00:50.44 | 645K| 31282 | | | | | 21 | UNION-ALL | | 1 | | 303 |00:00:50.44 | 645K| 31282 | | | | | 22 | VIEW | V_POSNR_0200011 | 1 | 2 | 20 |00:00:50.42 | 429K| 31244 | | | | | 23 | UNION-ALL | | 1 | | 20 |00:00:50.42 | 429K| 31244 | | | | | 24 | NESTED LOOPS | | 1 | 1 | 20 |00:00:50.42 | 376K| 28979 | | | | |* 25 | HASH JOIN | | 1 | 1 | 20 |00:00:50.42 | 376K| 28979 | 1096K| 1096K| 1348K (0)| |* 26 | TABLE ACCESS BY INDEX ROWID | PROPERTIES | 1 | 6 | 2651 |00:00:00.02 | 2131 | 0 | | | | |* 27 | INDEX RANGE SCAN | P_SETAALDATE_IDX | 1 | 6 | 2651 |00:00:00.01 | 21 | 0 | | | | | 28 | VIEW | VW_JF_SET$7992605D | 1 | 2 | 504 |00:02:30.85 | 374K| 28979 | | | | | 29 | UNION-ALL | | 1 | | 504 |00:02:30.85 | 374K| 28979 | | | |
I thought that the columns A-time indicates the time of the operation in question (summarizing all time from the operation of the child)
but this seems different:
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The Timing of the HASH (line 25) JOIN is 00:00:50.42, but the timing of the VIEW is 00:02:30.85 who ought both to complete before you chop it can happen
So I thought the HASH JOIN would be (at least) a schedule of the 00:02:30.85 of the VIEW, plus the 00:00:00.02 of the TABLE ACCESS BY INDEX ROWID.
But it seems that it is a misconception, can anyone shed light on this?
Best regards
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---------------------------------------------------------------------------------------------------
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8 selected lines
COUNT (*)
----------------------
7529536
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----------------------------------------------------------
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-----------------------------------------------------------------------------
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-----------------------------------------------------------------------------
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----------------------------------------------------------
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0 redo size
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396 bytes received via SQL * Net from client
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Ausf³hrungsplan
----------------------------------------------------------
Hash value of plan: 870989070
-----------------------------------------------------------------------------
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-----------------------------------------------------------------------------
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display the same value in a times column in multiple lines
Dear all:
I have a table id, with the following data values
ID VALUES
===============
EG001 5888314
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===============
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Hi Experts,
I had a doubt about the Plan to explain it.
Is there a relationship between the cost amount in the Plan to explain and the number of lines read by the query?
In other words - if the number of lines read by the query increases, which will increase the cost also?
Can someone please give me some advice on this?
Thank you and best regards,
Vanessa B.ranitB wrote:
Hi Experts,I had a doubt about the Plan to explain it.
Is there a relationship between the cost amount in the Plan to explain and the number of lines read by the query?
In other words - if the number of lines read by the query increases, which will increase the cost also?Can someone please give me some advice on this?
"The answer is 'not really' the cost is single block read disc of research time. But if the number of lines that are expected + coming in overall result are higher, the work expected by Oracle would be also higher. Thus, indirectly, the answer may be Yes. That said, the number of rows contained in the plan of the explain command (you should check the execution plan, explain plan is not the real plan) are the lines that the optimizer is expected to pick up and extracted from the number ofrows real may or may not match with her.
Aman...
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