Ethan,
Thanks for the alternative code.
I got the exact same time either way.
table "a" = 7000 records, table "b" = 30,000 records.
old netware/oracle7 server.
1 select
2 count( m.student_ssn )
3 from
4 student_master m,
5 SIS_CSUS_ALL_FALL_2001_eos1 x
6 where
7 m.student_ssn = x.stu_id (+)
8 and
9* x.stu_id is null
SQL> /
COUNT(M.STUDENT_SSN)
--------------------
5406
(35 seconds)
1 select
2 count( m.student_ssn )
3 from
4 student_master m
5 where
6 not exists
7 (
8 select
9 null
10 from
11 SIS_CSUS_ALL_FALL_2001_eos1 x
12 where
13 m.student_ssn = x.stu_id
14* )
SQL> /
COUNT(M.STUDENT_SSN)
--------------------
5406
(35 seconds)
best regards,
ep
On 23 Jan 2002 at 1:05, Oracle RDBMS Community Forum
wrote:
ORACLE-L Digest -- Volume 2002, Number 023
> ------------------------------
>
> From: "Post, Ethan" <[EMAIL PROTECTED]>
> Date: Tue, 22 Jan 2002 17:17:23 -0600
> Subject: RE: Sql query
>
> I believe this will work also and may be faster...
>
> select c1 from t1 a where not exists (select null
> from t2 b where a.c1 = b.c1);
>
>
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