崗位與候選人匹配查詢 - Job id and skill set query

Candidates

CandidateId	Skill
1	Sql
1	Dw
1	ssis
2	ssis
2	sql
2	vb

Jobs

JobId	SkillReq
3	Sql
3	Dw
4	ssis
4	sql
4	vb

1. 一條 sql 查出所有符合崗位要求的候選人（候選人技能包括崗位技能要求），即所有 JobId-CandidateId 對。
2. 一條 sql 查出所有精確符合崗位要求的候選人（候選人技能等於崗位技能要求）

Expected output would be, Job_ID need the respective candidate ID with exact skill set match.

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建表

CREATE TABLE Candidates (CandidateId INT,  Skill VARCHAR(100));
INSERT INTO Candidates VALUES (1,'Sql'), (1,'Dw'), (1,'ssis'), (2, 'ssis'), (2,'sql'), (2,'vb'); 

CREATE TABLE Jobs (JobId INT, SkillReq VARCHAR(100));
INSERT INTO Jobs VALUES (3,'Sql'), (3,'Dw'), (4,'ssis'), (4,'sql'), (4,'vb');

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1. 模糊查詢（Candidate.Skill_set >= JobId.SkillReq_set）

SELECT a.jid AS JobId, a.cid AS CandidateId FROM
	(SELECT J.JobId AS jid, C.CandidateId AS cid, count(1) AS count FROM Jobs J, Candidates C WHERE	J.SkillReq=C.Skill GROUP BY J.JobId, C.CandidateId) AS a
INNER JOIN 
	(SELECT JobId AS jid, count(1) AS count FROM Jobs GROUP BY JobId) AS b
ON a.jid = b.jid AND a.count = b.count;

• 先對兩個表做有條件的笛卡爾積，爲子句 a

比如集合A={a, b}, 集合B={0, 1, 2}，則二者的笛卡爾積爲 [(a,0), (a,1), (a,2), (b,0), (b,1), (b,2)]

SELECT J.JobId AS jid, C.CandidateId AS cid, count(1) AS count FROM Jobs J, Candidates C GROUP BY J.JobId, C.CandidateId

舉個例子，JobId=3 包括（Sql, Dw），CandidateId=1 包括（Sql, Dw, ssis），兩者的笛卡爾積應該是： (Sql, Sql), (Sql, Dw), (Sql,ssis)； (Dw, Sql), (Dw, Dw), (Dw, ssis)。

總計數量是6，爲了得到技能匹配的數量2，須加限定條件：

SELECT J.JobId AS jid, C.CandidateId AS cid, count(1) AS count FROM Jobs J, Candidates C WHERE	J.SkillReq=C.Skill GROUP BY J.JobId, C.CandidateId

此子句將得到技能匹配的情況，：

JobId	CandidateId	skill
3	1	Sql
3	1	Dw
3	2	Sql
4	1	ssis
4	1	Sql
4	2	ssis
4	2	vb

我們可以據此得到對應的匹配數量，比如 3-1:2， 3-2:1， 4-1:2， 4-2:2

• 查詢 Jobs 表，得到每個崗位要求的技能數量，爲子句 b

SELECT JobId AS jid, count(1) AS count FROM Jobs GROUP BY JobId；

• 基於以上兩個子句，做 INNER JOIN， 條件是 jobid 相同 且 數量相同。

SELECT a.jid AS JobId, a.cid AS CandidateId FROM
	(SELECT J.JobId AS jid, C.CandidateId AS cid, count(1) AS count FROM Jobs J, Candidates C WHERE	J.SkillReq=C.Skill GROUP BY J.JobId, C.CandidateId) AS a
INNER JOIN 
	(SELECT JobId AS jid, count(1) AS count FROM Jobs GROUP BY JobId) AS b
ON a.jid = b.jid AND a.count = b.count;

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2. 精確查詢（Candidate.Skill_set = JobId.SkillReq_set）

• 將Jobs 表中每個 JobId 的 SkillReq 升序排列，然後組合成字符串, 形成子句 J

(SELECT JobId, LOWER(GROUP_CONCAT(Distinct SkillReq ORDER BY SkillReq ASC SEPARATOR ',')) AS JS
FROM Jobs GROUP BY JobId) 
AS J

• 將 Candidates 表中每個 CandidateId 對應的技能 Skill 升序 排列，然後組合成字符串, 形成子句 C：

(SELECT CandidateId, LOWER(GROUP_CONCAT(DISTINCT Skill ORDER BY Skill ASC SEPARATOR ',')) AS CS
FROM Candidates GROUP BY CandidateId) 
AS C

• 最後聯表查詢，條件是 字符串相等:

SELECT J.JobId, C.CandidateId FROM

(SELECT JobId, LOWER(GROUP_CONCAT(DISTINCT SkillReq ORDER BY SkillReq ASC SEPARATOR ',')) AS JS FROM Jobs GROUP BY JobId) AS J

INNER JOIN

(SELECT CandidateId, LOWER(GROUP_CONCAT(DISTINCT Skill ORDER BY Skill ASC SEPARATOR ',')) AS CS FROM Candidates GROUP BY CandidateId) AS C

ON J.JS = C.CS;

注：主要是用到了函數 group_concat； 另外，加 lower 函數的原因是，去除大小寫干擾。