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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