RE: Question on how cardinality is calculated.

From: Ronnie Doggart <>
Date: Thu, 10 Jan 2008 17:32:48 -0000
Message-ID: <>


Thanks to those who answered, the majority pointing out that it is 5% of the number of rows.  

The oracle version is  

To build on the information provided, the actual query used in an application does the following:  

select cases.a, cases.b, cases.c,, enquiry.a, enquiry.b, enquiry.c from cases, enquiry
where =
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: on behalf of Jonathan Lewis Sent: Thu 1/10/2008 4:22 PM
To: Ronnie Doggart; Subject: Re: Question on how cardinality is calculated.

5% of num_rows


Jonathan Lewis <>

Author: Cost Based Oracle: Fundamentals

The Co-operative Oracle Users' FAQ

  • Original Message ----- From: "Ronnie Doggart" <> To: <> 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%';

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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-- Received on Thu Jan 10 2008 - 11:32:48 CST

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