RE: Question on how cardinality is calculated.
Date: Thu, 10 Jan 2008 17:32:48 -0000
Thanks to those who answered, the majority pointing out that it is 5% of the number of rows.
The oracle version is 184.108.40.206
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
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.
From: oracle-l-bounce_at_freelists.org on behalf of Jonathan Lewis
Sent: Thu 1/10/2008 4:22 PM
To: Ronnie Doggart; oracle-l_at_freelists.org Subject: Re: Question on how cardinality is calculated.
5% of num_rows
Author: Cost Based Oracle: Fundamentals
The Co-operative Oracle Users' FAQ
- Original Message ----- From: "Ronnie Doggart" <ronnie_doggart_at_lagan.com> To: <oracle-l_at_freelists.org> Sent: Thursday, January 10, 2008 3:06 PM Subject: Question on how cardinality is calculated.
Does anyone know how Oracle calculates the cardinality of a query such as:
Select * from table_1 where upper(case) like '%12345%';
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 ?
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http://www.freelists.org/webpage/oracle-l Received on Thu Jan 10 2008 - 11:32:48 CST