PLAN (B_TABLE NATURAL)
PLAN (AT INDEX (IDX_A_TABLE))

Does this mean that for every single row selected from table A, it
touches every single row from table B?

-----Original Message-----
From: firebird-support@yahoogroups.com
[mailto:firebird-support@yahoogroups.com] On Behalf Of Leyne, Sean
Sent: Monday, April 02, 2012 5:13 PM
To: firebird-support@yahoogroups.com
Subject: RE: [firebird-support] Not Exists cost

> If B_TABLE.VALUE2 never contains 'X' and half of A_TABLE.VALUE1 are

'Y'

> what would be the cost of executing this statement? A_TABLE.VALUE1 is

> indexed and B_TABLE.VALUE2 does not have an index.
>
> Would it be half of A_TABLE rows times all of B_TABLE rows? Or would
> the Not Exists statement be evaluated for every row in A_TABLE making
> it A_TABLE rows times B_TABLE rows?
>
> SELECT *
> FROM A_TABLE at
> WHERE
> at.VALUE1 = 'Y' AND
> NOT EXISTS (
> SELECT 1
> FROM B_TABLE bt
> WHERE bt.VALUE2 = 'X');

The query PLAN would tell you for certain, but I would expect the answer
to be:
half of A_TABLE rows times all of B_TABLE rows.


Sean


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