A script for Elland's sake. I hope it helps :D

Jobs_Skills_and_Candidates.sql

/**
* This sh*t is a dummy query to help a jenius fréndo
*
* The problem: get a dataset with the Job vs Candidate - Skill analysis, and answer: which candidate better fits the given job skills?
*
* Consider: 
* - A Job is composed by many Skills
* - And so is the Candidate...
*
* Use your imagination, cuz this is just a dummy script, I wont create any schema... >:D
*/

select
	-- you'd better see the "from clause" before you read the "select clause"...
	jb.job_id, -- id, description, whatever!
	cn.candidate_id, -- id, name, whatever!
	-- after you have defined your dataset (that's why I asked you to read "from" before), we will "count as many matches the condidate has"...
	-- 
	sum(
		case 
			-- the trick is simple: when it matches, it isnt null... so we count up once
			when skc.skill_id is not null then 1
			-- ... because of the left join, you have this effect, ta-da!
			else 0
		end
	) / -- ... and divede it by the number of "matches it could have", that is drive by the "Logical Job dataset". We are done!
	count(*) as percent_of_aproval
from
	-- assuming that your candidate has many skills and so the skills cloud be used by many candidates. The skill is the "strong entity", because you "always have a job" but the same isn't true for the candidate, isn't it right? (otherwise, we wouldnt have to do the SQL...)
	Job jb
	inner join JobSkill jbs 
		on jbs.job_id = jb.job_id
	inner join Skill skj -- those are trully the job skills
		on skj.skill_id = jbs.skill_id
	-- ok, so far, we have a "Logical job with its skills"
	-- now, to let's assemble the "Logical candidate blablablah"
	-- I'm using the parenthesis for a better readability and acurrate definition of our dataset
	left join (
		Candidate cn
		inner join CandidateSkill cns -- see? using parenthesis, I have full controll of join precedences!
			on cns.candidate_id = cn.candidate_id
		inner join Skill skc -- weird? np, we indeed need the "twice-joined" table... dont worry, it is fine =D (and still, it's very fast)
			on skc.skill_id = cns.skill_id
	) -- note: this isn't a "sub query", thus, CAN'T have any aliases
		-- now, the trick:
		-- we need to match "both logical entities". And ye know what ye need, dontchya?
		on skj.skill_id = skc.skill_id
where
	jb.job_id = {your_filter}
group by
	jb.job_id, -- id, description, whatever!
	cn.candidate_id; -- id, name, whatever!