Ian,
You just had to throw down the gauntlet, didn't you? :)
Here's my solution. You may consider it cheating, as I've done pretty much
what your
developer came up with, but I used a stored funtion.
This can probably be done in 'pure' SQL, but I don't have enough time to
ponder it right now.
Incidentally, I concatenate a ':' in the string with every course id?
Guess why? ;)
Jared
---- create the test table
drop table ian;
create table ian ( emp_id number(2), course_id number(2));
insert into ian values( 1, 1 );
insert into ian values( 2, 2 );
insert into ian values( 2, 3 );
insert into ian values( 3, 3 );
insert into ian values( 3, 4 );
insert into ian values( 3, 5 );
insert into ian values( 4, 3 );
insert into ian values( 4, 4 );
insert into ian values( 4, 5 );
insert into ian values( 5, 2 );
insert into ian values( 5, 3 );
insert into ian values( 6, 1 );
insert into ian values( 7, 2 );
insert into ian values( 8, 3 );
insert into ian values( 8, 4 );
commit;
--- create the function
create or replace function get_course_set( emp_id_in ian.emp_id%type )
return varchar2
is
vCourseStr varchar2(100);
begin
for emprec in (
select course_id
from ian
where emp_id = emp_id_in
) loop
vCourseStr := vCourseStr || to_char(emprec.course_id) || ':' ;
end loop;
return vCourseStr;
end;
/
--- run the test
col course_set format a30 head 'COURSE SET'
break on course_set skip 1
select
get_course_set(emp_id) course_set
, emp_id
from ian
group by emp_id
order by 1
/
"MacGregor,
Ian A." To: Multiple recipients of list ORACLE-L
<[EMAIL PROTECTED]>
<[EMAIL PROTECTED] cc:
ford.EDU> Subject: RE: An Interesting Grouping
Question | One Solution
Sent by:
[EMAIL PROTECTED]
om
08/31/01 11:00
AM
Please respond
to ORACLE-L
I could not think of a way to do it with any of the analytical functions.
The developer's method was to make an array with
one element being the emplid and the other a string with all that employees
courses sorted and concatenated together. Then one can group the
employees by comparing the strings.
There are many superb SQL programmers on the list. I thought one might
take a stab at it.
Ian MacGregor
Stanford Linear Accelerator Center
[EMAIL PROTECTED]
-----Original Message-----
Sent: Thursday, August 30, 2001 2:47 PM
To: Multiple recipients of list ORACLE-L
"MacGregor, Ian A." wrote:
>
> Given the following Table
>
> emplid course_id
> ------ ---------
> 1 1
> 2 2
> 2 3
> 3 3
> 3 4
> 3 5
> 4 3
> 4 4
> 4 5
> 5 2
> 5 3
> 6 1
> 7 2
> 8 3
> 8 4
>
--------------------------------------------------------------------------------------------
> What statement would you write to group employees by the set of courses
they have taken. In otherwords each employee in a group must have taken
the same as the others in the group, not one class more nor less. In this
example the employees making up the groups would be
>
> 1,6
> 2,5
> 3,4
> 7
> and 8
>
> I had this posed by one of my developers. He had also come up with a
solution which didn't take a relational approach. The approach is not
exotic, and I suspect it will be proposed by many people. He'd like a
relational one.
>
> Ian MacGregor
> Stanford Linear Accelerator Center
> [EMAIL PROTECTED]
I don't think you can, unless, perhaps, you use the analytical functions
which I have never had the opportunity to play with. Problem number 1
is getting an identifier for each set of courses. Since you must get
this through a 'GROUP BY', the only chance is a numerical expression. An
obvious candidate is something like sum(course_id * power(10, n -1))
where n is the order (starting with 1) of course ids suitably ordered
for each employee - restarting from 1 with each employee. Getting n is
the trouble. You cannot get it through rownums and in-line views, it
would require some kind of ugly three-way correlation between views in
the FROM clause and a subquery. The best solution I see, but it's not a
'pure play' one, is to create a function
create or replace function course_index(p_emplid in number,
p_course_id in number)
return number
as
n_val number;
begin
select a.n
into n_val
from (select b.course_id, rownum n
from (select course_id
from courses_taken
where emplid = p_emplid
order by 1 desc) b) a
where a.course_id = p_course_id;
return n
exception
when no_data_found then
return null;
end;
Then it becomes relatively easy to write
select emplid, sum(course_id * power(10, course_index(emplid,
course_id) - 1) course_set
from courses_taken
group by emplid
and then to do whatever you want.
--
Regards,
Stephane Faroult
Oriole Corporation
Voice: +44 (0) 7050-696-269
Fax: +44 (0) 7050-696-449
Performance Tools & Free Scripts
--------------------------------------------------------------
http://www.oriole.com, designed by Oracle DBAs for Oracle DBAs
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