GROUP BY and HAVING can appear in either order.

Hi,
please give me one example for the following line i found it in "19-28 Oracle Database SQL Language Reference".
If you specify both GROUP BY and HAVING, then they can appear in
either order.
yours sincerly

Hi,
944768 wrote:
that means both queries bring same result?Right; both ways produce the same results.
GROUP BY deptno
HAVING COUNT (*) > 2does exactly the same thing, just as efficiently as
HAVING COUNT (*) > 2
GROUP BY deptno

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Group by and having

Hi,
According to standard sql, we must have:
select ___
from ___
where ___
group by ___
having ___;
However, I tried do the query with the following order:
select ___
from ___
where ___
having ___
group by ___;
And I received the same result.
Does anybody knows the difference between those two orders? Are they always give the same result?
Thank You

According to the 9i documentation:
Specify GROUP BY and HAVING after the where_clause and CONNECT BY clause. If you specify both GROUP BY and HAVING, they can appear in either order.
Which you've got to admit does match your experience.
Cheers, APC

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Use of group by and having clause

hi frnds
can anybody explain me the use of group by an having clause in select state ment

Hi Rohit,
1. GROUP BY f1 ... fn
2. GROUP BY (itab)
1. GROUP BY f1 ... fn
Effect
Groups database table data in a SELECT command on one line in the result set. A group is a set of lines which all have the same values in each column determined by the field descriptors f1 ... fn.
... GROUP BY f1 ... fn always requires a list in the SELECT clause. If you use field descriptors without an aggregate funciton in the SELECT clause, you must list them in the GROUP BY f1 ... fn clause.
Example
Output the number of passengers, the total weight and the average weight of luggage for all Lufthansa flights on 28.02.1995:
TABLES SBOOK.
DATA: COUNT TYPE I, SUM TYPE P DECIMALS 2, AVG TYPE F.
DATA: CONNID LIKE SBOOK-CONNID.
SELECT CONNID COUNT( * ) SUM( LUGGWEIGHT ) AVG( LUGGWEIGHT )
       INTO (CONNID, COUNT, SUM, AVG)
       FROM SBOOK
       WHERE
         CARRID   = 'LH'       AND
         FLDATE   = '19950228'
       GROUP BY CONNID.
WRITE: / CONNID, COUNT, SUM, AVG.
ENDSELECT.
Note
... GROUP BY f1 ... fn is not supported for pooled and cluster tables.
2. GROUP BY (itab)
Effect
Works like GROUP BY f1 ... fn if the internal table itab contains the list f1 ... fn as ABAP source code. The internal table itab can only have one field. This field must be of the type C and should not be more than 72 characters long. itab must be enclosed in parentheses and there should be no blanks between the parentheses and the table name.
Note
The same restrictions apply to this variant as to GROUP BY f1 ... fn.
Example
Output all Lufthansa departure points with the number of destinations:
TABLES: SPFLI.
DATA:   BEGIN OF WA.
          INCLUDE STRUCTURE SPFLI.
DATA:     COUNT TYPE I.
DATA:   END OF WA.
DATA:   WA_TAB(72) TYPE C,
        GTAB LIKE TABLE OF WA_TAB,
        FTAB LIKE TABLE OF WA_TAB,
        COUNT TYPE I.
CLEAR: GTAB, FTAB.
WA_TAB = 'COTYFROM COUNT( * ) AS COUNT'. APPEND FTAB.
APPEND WA_TAB TO FTAB.
WA_TAB = 'CITYFROM'.
APPEND WA_TAB TO GTAB.
SELECT DISTINCT (FTAB)
       INTO CORRESPONDING FIELDS OF WA
       FROM SPFLI
       WHERE
         CARRID   = 'LH'
       GROUP BY (GTAB).
WRITE: / WA-CITYFROM, WA-COUNT.
ENDSELECT.
Regards,
Susmitha

Question about GROUP BY and HAVING

Good afternoon,
I have the following query which returns the desired result (set of students who take CS112 or CS114 but not both). I wanted to "condense" it into a single SELECT statement (if that is at all possible - DDL to execute the statement is provided at the end of this post):
-- is this select distinct * and its associated where clause absolutely
-- necessary to obtain the result ?
select distinct *
from (
        select s.sno,
               s.sname,
               s.age,
               sum(case when t.cno in ('CS112', 'CS114')
                        then 1
                        else 0
                   end)
                 over (partition by s.sno) as takes_either_or_both
          from student s join take t
                           on (s.sno = t.sno)
where takes_either_or_both = 1
;The following looked reasonable but, unfortunately unsuccessful:
Window functions not allowed here (in Having Clause)
select max(s.sno),
       max(s.sname),
       max(s.age),
       sum(case when t.cno in ('CS112', 'CS114')
                then 1
                else 0
           end)
         over (partition by s.sno) as takes_either_or_both
from student s join take t
                   on (s.sno = t.sno)
group by s.sno
having sum(case when t.cno in ('CS112', 'CS114')
                then 1
                else 0
           end)
         over (partition by s.sno) = 1
Invalid identifier in Having clause
select s.sno,
       s.sname,
       s.age,
       sum(case when t.cno in ('CS112', 'CS114')
                then 1
                else 0
           end)
         over (partition by s.sno) as takes_either_or_both
from student s join take t
                   on (s.sno = t.sno)
group by s.sno, s.sname, s.age
having takes_either_or_both = 1
;I have searched for a document that completely defines the sequence in which the clauses are executed. I have found tidbits here and there but not something complete. I realize that my running into problems like this one is due to my lack of understanding of the sequence and the scope of the clauses that make up a statement. Because of this, I cannot even tell if it is possible to write the above query using a single select statement. Pardon my bit of frustration...
Thank you for your help,
John.
DDL follows.
        /* drop any preexisting tables */
        drop table student;
        drop table courses;
        drop table take;
        /* table of students */
        create table student
        ( sno integer,
          sname varchar(10),
          age integer
        /* table of courses */
        create table courses
        ( cno varchar(5),
          title varchar(10),
          credits integer
        /* table of students and the courses they take */
        create table take
        ( sno integer,
          cno varchar(5)
        insert into student values (1,'AARON',20);
        insert into student values (2,'CHUCK',21);
        insert into student values (3,'DOUG',20);
        insert into student values (4,'MAGGIE',19);
        insert into student values (5,'STEVE',22);
        insert into student values (6,'JING',18);
        insert into student values (7,'BRIAN',21);
        insert into student values (8,'KAY',20);
        insert into student values (9,'GILLIAN',20);
        insert into student values (10,'CHAD',21);
        insert into courses values ('CS112','PHYSICS',4);
        insert into courses values ('CS113','CALCULUS',4);
        insert into courses values ('CS114','HISTORY',4);
        insert into take values (1,'CS112');
        insert into take values (1,'CS113');
        insert into take values (1,'CS114');
        insert into take values (2,'CS112');
        insert into take values (3,'CS112');
        insert into take values (3,'CS114');
        insert into take values (4,'CS112');
        insert into take values (4,'CS113');
        insert into take values (5,'CS113');
        insert into take values (6,'CS113');
        insert into take values (6,'CS114');

Hi, John,
Just use the aggregate SUM function.
        select s.sno,
               s.sname,
               s.age,
               sum(case when t.cno in ('CS112', 'CS114')
                        then 1
                        else 0
                   end) as takes_either_or_both
          from student s join take t
                           on (s.sno = t.sno)
       GROUP BY s.sno,
                s.sname,
              s.age
       HAVING sum(case when t.cno in ('CS112', 'CS114')
                        then 1
                        else 0
                   end) = 1;Analytic functions are computed after the WHERE- and HAVING clause have been applied. To use the results of an analytic fucntion in a WHERE- or HAVING clause, you have to compute it in a sub-query, and then you can use it in a WHERE- or HAVING clause of a super-query.

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Find duplicate records withouyqusing group by and having

I know i can delete duplicate records using an analytic function. I don't want to delete the records, I want to look at the data to try to understand why I have duplicates. I am looking at tables that don't have unique constraints (I can't do anything about it). I have some very large tables. so I am trying to find a faster way to do this.
for example
myTable
col1 number,
col2 number,
col3 number,
col4 number,
col5 number)
My key column is col1 and col2 (it is not enforced in the database). So I want to get all the records that have duplicates on these fields and put them in a table to review. This is a standard way to do it, but it requires 2 full table scans of very large tables (many, many gigabtytes. one is 150 gbs and not partitioned), a sort, and a hash join. Even if i increase sort_area_size and hash_area_size, it takes a long time to run..
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as
select b.*
from (select col1,col2,count(*)
from myTable
group by col1, col2
having count(*) > 1) a,
myTable b
where a.col1 = b.col1
and a.col2 = b.col2
I think there is a way to do this without a join by using rank, dense_rank, or row_number or some other way. When I google this all I get is how to "delete them". I want to analyze them a nd not delete them.

create table mytable (col1 number,col2 number,col3 number,col4 number,col5 number);
insert into mytable values (1,2,3,4,5);
insert into mytable values (2,2,3,4,5);
insert into mytable values (3,2,3,4,5);
insert into mytable values (2,2,3,4,5);
insert into mytable values (1,2,3,4,5);
SQL> ed
Wrote file afiedt.buf
1 select * from mytable
2   where rowid in
3 (select rid
4      from
5     (select rowid rid,
6              row_number() over
7              (partition by
8                   col1,col2
9               order by rowid) rn
10          from mytable
11      )
12    where rn <> 1
13* )
SQL> /
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         1          2          3          4          5
         2          2          3          4          5
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GROUP BY and HAVING can appear in either order.

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