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RE: An Interesting Grouping Question | One Solution

From: MacGregor, Ian A. <ian_at_SLAC.Stanford.EDU>
Date: Fri, 31 Aug 2001 09:55:47 -0700
Message-ID: <F001.0037FA0C.20010831100017@fatcity.com>

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
ian_at_slac.stanford.edu

-----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
> ian_at_slac.stanford.edu

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 
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Author: MacGregor, Ian A.
  INET: ian_at_SLAC.Stanford.EDU

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Received on Fri Aug 31 2001 - 11:55:47 CDT

Original text of this message

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