RE: Question on how cardinality is calculated.
- From: "Elliott, Patrick" <patrick.elliott@xxxxxxxxxxxxx>
- To: "ronnie_doggart@xxxxxxxxx" <ronnie_doggart@xxxxxxxxx>, "jonathan@xxxxxxxxxxxxxxxxxx" <jonathan@xxxxxxxxxxxxxxxxxx>, "oracle-l@xxxxxxxxxxxxx" <oracle-l@xxxxxxxxxxxxx>
- Date: Thu, 10 Jan 2008 12:47:40 -0600
I hope you have a function based index on upper(case). If so, then you could
try an index hint or first_rows hint, but it will still do a full index scan
because the "like '%123456%'" can't be done with a range scan. Why do you need
to do an upper if you are going to have the user entering digits anyway?
Upper('123456') is the same as '123456'. Take the upper out and then you can
create a regular index. That could be the whole problem to begin with. A
regular index will not be used if you have a function on the column.
Pat
-----Original Message-----
From: oracle-l-bounce@xxxxxxxxxxxxx [mailto:oracle-l-bounce@xxxxxxxxxxxxx] On
Behalf Of Ronnie Doggart
Sent: Thursday, January 10, 2008 11:33 AM
To: jonathan@xxxxxxxxxxxxxxxxxx; oracle-l@xxxxxxxxxxxxx
Subject: RE: Question on how cardinality is calculated.
All,
Thanks to those who answered, the majority pointing out that it is 5% of the
number of rows.
The oracle version is 9.2.0.8
To build on the information provided, the actual query used in an application
does the following:
select cases.a, cases.b, cases.c, cases.case, enquiry.a, enquiry.b, enquiry..c
from cases, enquiry where enquiry.id = cases.id and upper(case) like '%123456%';
Now the user always enters six digits which I know identifies 1 record in the
cases table, but because Oracle uses a cardinality of 5553 for this it performs
a hash join to the enquiry table, even though only one record will match in the
enquiry table. The enquiry table contains 1.4 million record so it takes ~5
seconds.
Would it be advisable to put a hint on the query to either use a nested loop
which allows the query to use an index to retrieve the data from enquiry table.
Or use a cardinality hint since I know we will not be returning 5553 rows which
will also allow the use of an index.
Thanks
Ronnie
________________________________
From: oracle-l-bounce@xxxxxxxxxxxxx on behalf of Jonathan Lewis
Sent: Thu 1/10/2008 4:22 PM
To: Ronnie Doggart; oracle-l@xxxxxxxxxxxxx
Subject: Re: Question on how cardinality is calculated.
5% of num_rows
Regards
Jonathan Lewis
http://jonathanlewis.wordpress.com <http://jonathanlewis.wordpress.com/>
Author: Cost Based Oracle: Fundamentals
http://www.jlcomp.demon.co.uk/cbo_book/ind_book.html
The Co-operative Oracle Users' FAQ
http://www.jlcomp.demon.co.uk/faq/ind_faq.html
----- Original Message -----
From: "Ronnie Doggart" <ronnie_doggart@xxxxxxxxx>
To: <oracle-l@xxxxxxxxxxxxx>
Sent: Thursday, January 10, 2008 3:06 PM
Subject: Question on how cardinality is calculated.
All
Does anyone know how Oracle calculates the cardinality of a query such as:
Select * from table_1 where upper(case) like '%12345%';
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=131 Card=5553 Bytes=621936)
1 0 TABLE ACCESS (FULL) OF 'TABLE_1' (Cost=131 Card=5553 Bytes=621936)
Oracle calculates that it is expecting 5553 rows from this query but I know it
will only return 2.
How does it do its estimate ?
Ronnie Doggart
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