Re: Performance off "count(*)"

  • From: "Jonathan Lewis" <jonathan@xxxxxxxxxxxxxxxxxx>
  • To: <oracle-l@xxxxxxxxxxxxx>
  • Date: Fri, 18 Jul 2008 15:20:11 +0100

In principle the optimizer knows that count(*) means
"how many rows in the table" and will choose the smallest object that could return the result to do the scan. This means it may choose to do a fast-full scan of an index if at least one column of the index is declared
as not null.

Depending on  version, the choice between indexes of
identical cost may simply be based on alphabetical ordering
of the names - although number of distinct keys is supposed to have an effect as well in recent versions


Jonathan Lewis

Author: Cost Based Oracle: Fundamentals

The Co-operative Oracle Users' FAQ

----- Original Message ----- From: "Marco Gralike" <Marco.Gralike@xxxxxxx>
To: "Gints Plivna" <gints.plivna@xxxxxxxxx>
Cc: <oracle-l@xxxxxxxxxxxxx>
Sent: Friday, July 18, 2008 2:11 PM
Subject: RE: Performance off "count(*)"

That's not really what I ment.

While using "count(*)", Oracle will performance optimize (as far as I
know) the "count(*)" so it will do a smarter job then "only" a full
table scan.

I am interested in the how and what regarding the mechanics / methods
behind it. In my "not so relational XMLDB" environment, I see a full
table scan were I would have hoped for the smarter way to go for an
index. The count via the index will last for 3 minutes. The count via
the full table scan will take more than 1 day (17 Gb of XML data,
approx. 7 milion records).

I have an small example of this smarter behavior below (I thought it was
somehow hardcoded):

Also see the differences (I have to force it to go via the PK unique
index via a hint)


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