Re: Select count(*)

Hi Tom,

   I´m not using both tables in queries, I´m using one table and comparing the execution time with the other table. The SQL´s are the same, the only difference is the table in the FROM clause.

Regards,
Carlos

Thomas Kyte <tkyte_at_oracle.com> wrote in message news:<ag6qsg02gau_at_drn.newsguy.com>...
> In article <72954535.0207051011.1058d2be_at_posting.google.com>,
> calberto2312_at_hotmail.com says...
> >
> >Hi,
> >
> > I have two tables (A and B), with the same structure, but in
> >different tablespaces. One (A) has 50 million records and was created
> >by the other one (B) (by INSERT ... SELECT ....), with 100 million
> >records. After the data load, the indexes were created and 'analyze
> >compute' executed. Then, after I made some performance tests, the
> >results seem very strange :
> >
> >Test 1 :
> >
> >set timing on;
> >
> >select <columns>, ....
> >from A, B
> >where <indexed_column> = ....
> >
> >Table B is faster, although it has the double of records. As more
> >times this SQL is executed, the time of execution of both tables are
> >almost the same.
> >
> >
> >Test 2 :
> >
> >set timing on;
> >
> >select count(*)
> >from A, B
> >where <indexed_column> = ....
> >
> >Table A is faster.
> >
> >
> >Test 3 :
> >
> >set timing on;
> >
> >select count(*) from A, B
> >
> >Table A is faster.
> >
> >
> > Why table A is just faster when I run select count(*)? I´m using
> >Oracle 8.0.6.0.0.
> >
> >Thanks in advanced,
> >Carlos
>
>
> I have no idea what you might mean by "table A is faster" or "table B is faster"
> since you use both A and B in all queries. Makes no sense....
Received on Mon Jul 08 2002 - 09:54:54 CDT