With Postgres 8.4, there are the new window functions

http://www.postgresql.org/docs/8.4/static/tutorial-window.html

That is if switching databases is an option

On Tue, Aug 17, 2010 at 9:33 AM, Brendan Brink
<[email protected]>wrote:

> sorry about that, MYSQL5
>
> On Tue, Aug 17, 2010 at 9:31 AM, Neven MacEwan <[email protected]> wrote:
> > What SQL? It makes a huge difference
> >
> >> hi there,
> >>
> >> have a query with building an SQL query...
> >>
> >> SELECT DISTINCT(e.id),cs2.company_id as companyid, count(es2.skill) as
> >> skillswanted FROM employees_personal e INNER JOIN employees_skills es2
> >> on es2.employee_id = e.id INNER JOIN company_skill cs2 on cs2.skill =
> >> es2.skill WHERE cs2.company_id = '117' AND 1 GROUP BY companyid order
> >> by skillswanted DESC, e.first_name,e.last_name
> >>
> >> Basically...
> >>
> >> Employees_personal have skills in the employees_skills table,
> >> Companies have skills they require in company_skill table
> >>
> >> i want to find all employees who have skills that match skills in the
> >> company skill table, and then order by the number of matches desc to
> >> find most suitable employee.
> >>
> >>
> >>
> >> ... currently the query is not giving any errors, but it only returns
> >> one employee, when I know there should be alot more results...
> >>
> >> also...ideally would like to create a HAVING clause on the number of
> >> skillswanted...so can say only display employees who match at least 2
> >> skills.
> >>
> >> thanks..for any help would be appreciated...
> >>
> >> --
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> >>
> >>
> >>
> >
> >
> > --
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> --
> Kind Regards,
>
> Brendan Brink
>
> SMS Marketing Consultant | Manager
> Sell2Cell Ltd.
>
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-- 
Best regards,
Y. Thong Kuah

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