Question:
[Joe Celko’s Employment Agency Puzzle]
You are running an employment agency that has requests for jobs and applications from candidates. Both of these forms include a list of the skills required or offered, and whether this is a primary or secondary skill. The tables look like this:
CREATE TABLE CandidateSkills (candidateid INTEGER NOT NULL PRIMARY KEY, name CHAR(20) NOT NULL, skillid INTEGER NOT NULL, skilllevel INTEGER NOT NULL CHECK (skilllevel IN (1, 2)), PRIMARY KEY (candidateid, skillid, skilllevel) ); CREATE TABLE JobSkills (jobid INTEGER NOT NULL, skillid INTEGER NOT NULL, skilllevel INTEGER NOT NULL CHECK (skilllevel IN (1, 2)), PRIMARY KEY (jobid, skillid, skilllevel) );We need to write some queries that will match up candidates with jobs. But our service offers three kinds of matchings for a job:
1. A list of candidate names who each exhibit ALL the primary skills and ANY of the secondary skills.
2. A list of candidate names who each exhibit ALL the primary skills and ALL of the secondary skills.
3. A list of candidate names who each exhibit ALL the primary skills and (n) number of the secondary skills.
Answer:
To find people with all the primary skills:
SELECT C1.name, C1.candidateid, J1.jobid FROM JobSkills AS J1 LEFT OUTER JOIN Candidates AS C1 ON J1.skilllevel = C1.skilllevel AND J1.skillid = C1.skillid; GROUP BY C1.name, C1.candidateid, J1.jobid HAVING ((SELECT COUNT(skillid) FROM Candidates AS C2 WHERE C1.candidateid = C2.candidateid AND C2.skilllevel = 1) >= (SELECT COUNT(skillid) FROM JobSkills AS J2 WHERE J1.jobid = J2.jobid AND J2.skilllevel = 1)) AND [add one of the following];a) All the secondary skills:
((SELECT COUNT(skillid) FROM Candidates AS C2 WHERE C1.candidateid = C2.candidateid AND C2.skilllevel = 2) >= (SELECT COUNT(skillid) FROM JobSkills AS J2 WHERE J1.jobid = J2.jobid AND J2.skilllevel = 2))b) Any of the secondary skills:
((SELECT COUNT(skillid) FROM Candidates AS C2 WHERE C1.candidateid = C2.candidateid AND C2.skilllevel = 2) > 0)c) exactly (n) of the secondary skills:
((SELECT COUNT(skillid) FROM Candidates AS C2 WHERE C1.candidateid = C2.candidateid AND C2.skilllevel = 2) = :n)
Puzzle provided courtesy of:
Joe Celko
[email protected]