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... >> >> -- >> NZ PHP Users Group: http://groups.google.com/group/nzphpug >> To post, send email to [email protected] >> To unsubscribe, send email to >> [email protected] >> >> >> > > > -- > NZ PHP Users Group: http://groups.google.com/group/nzphpug > To post, send email to [email protected] > To unsubscribe, send email to > [email protected]
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