how to match in mysql - mysql

Question

I have two tables, one for candidates and their skills, and the other for jobs and the skills required for the job, such as the ones shown below:
CSID = candidate skill ID (PK)
CID = candidate ID (FK)
S_CODE = skill code (FK)
Here is the candidate_skill table
+------------+---------+---------+
| CSID | CID | S_CODE |
+------------+---------+---------+
| 1 | 1 | 5 |
| 2 | 1 | 9 |
| 3 | 2 | 5 |
| 4 | 2 | 10 |
+------------+---------+---------+
SJID = skill job ID (PK)
JID = job ID (FK)
S_CODE = skill code (FK)
Here is the skill_job table:
+------------+---------+---------+
| SJID | JID | S_CODE |
+------------+---------+---------+
| 12 | 50 | 5 |
| 13 | 50 | 9 |
| 14 | 51 | 1 |
| 15 | 52 | 10 |
+------------+---------+---------+
So in this example, the only candidate would be 1 for the job 50, since the skill codes (S_CODE) are 5 and 9 for both, but i also would like candidate 2 to be a match to 52, since it has the required job skill and an extra one.
I have tried several ways to match the job and candidate but I'm failing, an example of such a way is the code below:
SELECT * , COUNT(skill_job.S_CODE) AS cnt FROM candidate_skill,
skill_job WHERE candidate_skill.S_CODE = skill_job.S_CODE HAVING (cnt >=2)
the problem is, that only limits the candidate found to one and if i remove the COUNT clause, it lists all candidates that match with only one skill so candidate 2 would also be matched to the job 50.
Here is the mysql code for the tables from phpmyadmin:
-- phpMyAdmin SQL Dump
-- version 3.4.5
-- http://www.phpmyadmin.net
--
-- Host: localhost
-- Generation Time: Jul 16, 2012 at 09:27 PM
-- Server version: 5.5.16
-- PHP Version: 5.3.8
SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO";
SET time_zone = "+00:00";
/*!40101 SET #OLD_CHARACTER_SET_CLIENT=##CHARACTER_SET_CLIENT */;
/*!40101 SET #OLD_CHARACTER_SET_RESULTS=##CHARACTER_SET_RESULTS */;
/*!40101 SET #OLD_COLLATION_CONNECTION=##COLLATION_CONNECTION */;
/*!40101 SET NAMES utf8 */;
--
-- Database: `employment`
--
-- --------------------------------------------------------
--
-- Table structure for table `candidate`
--
CREATE TABLE IF NOT EXISTS `candidate` (
`CID` int(4) NOT NULL AUTO_INCREMENT,
`title` varchar(5) NOT NULL,
`fname` varchar(30) NOT NULL,
`lname` varchar(30) NOT NULL,
`dob` date NOT NULL,
`email` varchar(50) NOT NULL,
`address` varchar(255) NOT NULL,
`city` varchar(50) NOT NULL,
`postcode` varchar(10) NOT NULL,
`phone_num` varchar(11) NOT NULL,
`username` varchar(40) NOT NULL,
`password` varchar(40) NOT NULL,
`regdate` datetime NOT NULL,
`acc_type` enum('c','s') NOT NULL DEFAULT 'c',
`emailactivate` enum('0','1') NOT NULL DEFAULT '0',
`cv_name` varchar(60) NOT NULL,
`cv` blob NOT NULL,
PRIMARY KEY (`CID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 COMMENT='// this is the table for the candidates' AUTO_INCREMENT=175 ;
--
-- Dumping data for table `candidate`
--
INSERT INTO `candidate` (`CID`, `title`, `fname`, `lname`, `dob`, `email`, `address`, `city`, `postcode`, `phone_num`, `username`, `password`, `regdate`, `acc_type`, `emailactivate`, `cv_name`, `cv`) VALUES
(128, 'Mr', 'clement', 'Chilingulo', '0000-00-00', 'chlngl#yahoo.com', '28 oakfield Road', 'london', 'E6 1LW', '07771611873', 'casante', 'b59c67bf196a4758191e42f76670ceba', '0000-00-00 00:00:00', 'c', '0', '', '');
INSERT INTO `candidate` (`CID`, `title`, `fname`, `lname`, `dob`, `email`, `address`, `city`, `postcode`, `phone_num`, `username`, `password`, `regdate`, `acc_type`, `emailactivate`, `cv_name`, `cv`) VALUES
(134, 'Mr', 'rverv', 'revrb', '0000-00-00', 'tdbsdrt', 'trsbrtd', 'trbtrtrb', 'tbrfbgrts', 'trfbtrgbrfg', 'clement', 'b59c67bf196a4758191e42f76670ceba', '0000-00-00 00:00:00', 's', '0', '', '');
INSERT INTO `candidate` (`CID`, `title`, `fname`, `lname`, `dob`, `email`, `address`, `city`, `postcode`, `phone_num`, `username`, `password`, `regdate`, `acc_type`, `emailactivate`, `cv_name`, `cv`) VALUES
(165, 'Mr', 'oinInINOioni', 'ioin', '0000-00-00', 'inioimn', 'in', 'oin', 'oni', 'io', 'k', '7b8b965ad4bca0e41ab51de7b31363a1', '0000-00-00 00:00:00', 'c', '0', '', ''),
(166, 'Mr', 'pjINoNlkinoinoi', 'ino', '0000-00-00', 'oimpnponi', 'inoi', 'no', 'nj', 'nio', 'nio', 'eb5bc837d01b911029ae378e8a1c9f5d', '0000-00-00 00:00:00', 'c', '0', '', ''),
(167, 'Mr', 'vrae', 'ergvaer', '0000-00-00', 'aerbg', 'aergvera', 'aergvrea', 'aergvear', 'aergarev', 'grebvarvf', '609c1b136ec8d0d2dfdf9a2105fb605f', '0000-00-00 00:00:00', 'c', '0', '', ''),
(168, 'Mr', 'vrae', 'ergvaer', '0000-00-00', 'aerbg', 'aergvera', 'aergvrea', 'aergvear', 'aergarev', 'grebvarvf', '609c1b136ec8d0d2dfdf9a2105fb605f', '0000-00-00 00:00:00', 'c', '0', '', ''),
(169, 'Mr', 'ubp', 'bu', '0000-00-00', 'ubip', ';ub', 'ubi', 'ubo', 'buo', 'ubiipbu', '9f44ce1389a3e7372834ed730b559a5e', '0000-00-00 00:00:00', 'c', '0', '', ''),
(170, 'Mr', 'rvev', 'ferhbgetb', '0000-00-00', 'tsbtrb', 'trbstrb', 'trbsfb ', 'stb stb', 'vvfs', 'csdcdsvarewdcv', 'd41d8cd98f00b204e9800998ecf8427e', '0000-00-00 00:00:00', 'c', '0', '', ''),
(171, 'Mr', 'rvev', 'ferhbgetb', '0000-00-00', 'tsbtrb', 'trbstrb', 'trbsfb ', 'stb stb', 'vvfs', 'csdcdsvarewdcv', 'd41d8cd98f00b204e9800998ecf8427e', '0000-00-00 00:00:00', 'c', '0', '', ''),
(172, '', '', '', '0000-00-00', '', '', '', '', '', '', 'd41d8cd98f00b204e9800998ecf8427e', '0000-00-00 00:00:00', 'c', '0', '', ''),
(173, '', 'dds', 'vrv av', '0000-00-00', 'rvear', 'vreverav', 'rvedfsdv', 'frvdvf', 'fveavd', 'vdrareavdv', '84c71ac76340092398a1ed4bc7d6fe19', '0000-00-00 00:00:00', 'c', '0', '', ''),
(174, '', 'dds', 'vrv av', '0000-00-00', 'rvedwear', 'vreverav', 'rvedfsdv', 'frvdvf', 'fveavd', 'vdraredfcavdv', 'd41d8cd98f00b204e9800998ecf8427e', '0000-00-00 00:00:00', 'c', '0', '', '');
-- --------------------------------------------------------
--
-- Table structure for table `candidate_skill`
--
CREATE TABLE IF NOT EXISTS `candidate_skill` (
`CSID` int(4) NOT NULL AUTO_INCREMENT,
`CID` int(4) NOT NULL,
`S_CODE` int(4) NOT NULL,
PRIMARY KEY (`CSID`),
KEY `CID` (`CID`),
KEY `S_CODE` (`S_CODE`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 COMMENT='//match candidate and skill' AUTO_INCREMENT=131 ;
--
-- Dumping data for table `candidate_skill`
--
INSERT INTO `candidate_skill` (`CSID`, `CID`, `S_CODE`) VALUES
(114, 134, 2),
(116, 134, 9),
(121, 134, 5),
(126, 128, 1);
-- --------------------------------------------------------
--
-- Table structure for table `job`
--
CREATE TABLE IF NOT EXISTS `job` (
`JID` int(4) NOT NULL AUTO_INCREMENT,
`job_title` varchar(40) NOT NULL,
`job_desc` varchar(255) NOT NULL,
`start_date` date NOT NULL,
`end_date` date NOT NULL,
PRIMARY KEY (`JID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 COMMENT='// this is the table for the job vacancies' AUTO_INCREMENT=10 ;
--
-- Dumping data for table `job`
--
INSERT INTO `job` (`JID`, `job_title`, `job_desc`, `start_date`, `end_date`) VALUES
(1, 'engineer', ' fix engineering ish that is messed ', '0000-00-00', '0000-00-00'),
(5, 'pilot', '', '0000-00-00', '0000-00-00'),
(8, 'Charity Helper', ' ', '0000-00-00', '0000-00-00'),
(9, 'accountant', ' ', '0000-00-00', '0000-00-00');
-- --------------------------------------------------------
--
-- Table structure for table `skill`
--
CREATE TABLE IF NOT EXISTS `skill` (
`S_CODE` int(4) NOT NULL AUTO_INCREMENT COMMENT '// this is the skill primary key',
`skill_name` varchar(40) NOT NULL,
`skill_desc` varchar(255) NOT NULL,
PRIMARY KEY (`S_CODE`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=11 ;
--
-- Dumping data for table `skill`
--
INSERT INTO `skill` (`S_CODE`, `skill_name`, `skill_desc`) VALUES
(1, 'speaking', 'English is your first language'),
(2, ' writing', 'You can write English very well'),
(3, ' public', ''),
(5, 'initiative', ''),
(6, 'interviewing', ''),
(7, 'negotiating', ''),
(8, 'leading', ''),
(9, ' energy', ''),
(10, ' organisation', '');
-- --------------------------------------------------------
--
-- Table structure for table `skill_job`
--
CREATE TABLE IF NOT EXISTS `skill_job` (
`SJID` int(4) NOT NULL AUTO_INCREMENT,
`JID` int(4) NOT NULL,
`S_CODE` int(4) NOT NULL,
PRIMARY KEY (`SJID`),
KEY `S_CODE` (`S_CODE`),
KEY `JID` (`JID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=106 ;
--
-- Dumping data for table `skill_job`
--
INSERT INTO `skill_job` (`SJID`, `JID`, `S_CODE`) VALUES
(91, 5, 2),
(94, 5, 5),
(95, 5, 9),
(98, 1, 1),
(102, 1, 8),
(105, 8, 8);
--
-- Constraints for dumped tables
--
--
-- Constraints for table `candidate_skill`
--
ALTER TABLE `candidate_skill`
ADD CONSTRAINT `candidate_skill_ibfk_3` FOREIGN KEY (`CID`) REFERENCES `candidate` (`CID`) ON DELETE CASCADE ON UPDATE CASCADE,
ADD CONSTRAINT `candidate_skill_ibfk_4` FOREIGN KEY (`S_CODE`) REFERENCES `skill` (`S_CODE`) ON DELETE CASCADE ON UPDATE CASCADE;
--
-- Constraints for table `skill_job`
--
ALTER TABLE `skill_job`
ADD CONSTRAINT `skill_job_ibfk_4` FOREIGN KEY (`JID`) REFERENCES `job` (`JID`) ON DELETE CASCADE ON UPDATE CASCADE,
ADD CONSTRAINT `skill_job_ibfk_5` FOREIGN KEY (`S_CODE`) REFERENCES `skill` (`S_CODE`) ON DELETE CASCADE ON UPDATE CASCADE;
/*!40101 SET CHARACTER_SET_CLIENT=#OLD_CHARACTER_SET_CLIENT */;
/*!40101 SET CHARACTER_SET_RESULTS=#OLD_CHARACTER_SET_RESULTS */;
/*!40101 SET COLLATION_CONNECTION=#OLD_COLLATION_CONNECTION */;

Answer 1

Double negation is your friend (though I don't know MySQL well enough to tell whether it supports it):
SELECT DISTINCT CS.CID, SJ.JID
FROM CANDIDATE_SKILL CS
JOIN SKILL_JOB SJ ON CS.S_CODE = SJ.S_CODE
WHERE NOT EXISTS(SELECT 1
FROM SKILL_JOB SJ2
WHERE SJ.JID = SJ2.JID
AND NOT EXISTS(SELECT 1
FROM CANDIDATE_SKILL CS2
WHERE CS.CID = CS2.CID
AND SJ2.S_CODE = CS2.S_CODE))
A 'human translation' of this is to say that there should not exists any skill required for the job that the candidate doesn't have.

Answer 2

It seems like you'll need to use an INNER JOIN to combine your two tables and then compare their row values.

Answer 3

Should be something like this..
SELECT
CID
FROM
candidate_skill
WHERE
S_CODE IN (
SELECT S_CODE FROM skill_jobs where JID = 52
)
GROUP BY CID
CORRECTION: Previous query will fail to match all required S_CODE
SELECT
CID, COUNT(CSID) as cnt
FROM
candidate_skill
INNER JOIN skill_jobs
WHERE
S_CODE IN (
SELECT S_CODE FROM skill_jobs where JID = 52
)
GROUP BY CID
HAVING cnt = (
SELECT count(*) from skill_jobs where JID = 52
)

Answer 4

You will need a bit more complicated JOIN, and I see no easy way of simultaneously answering your request and the obvious ones "How is candidate X matched for job Y?" and "HOW MUCH does candidate X fit job Y?".
Anyway, here goes.
SELECT candidate_skill.CID, skill_job.JID, COUNT(*) AS has, sjtot.needed
FROM skill_job JOIN candidate_skill ON (candidate_skill.S_CODE = skill_job.S_CODE)
LEFT JOIN ( SELECT JID, COUNT(*) AS needed FROM skill_job GROUP BY JID ) AS sjtot
ON ( skill_job.JID = sjtot.JID )
GROUP BY CID, JID
HAVING has >= needed;
In practice I first group the required skills on skill_job, and this tells me that job 50 requires two skills.
Armed with this, I LEFT JOIN required and candidate skills; some candidates will join on all skills, some won't. It is then just a matter of counting how many do match.
+------+------+-----+--------+
| CID | JID | has | needed |
+------+------+-----+--------+
| 1 | 50 | 2 | 2 |
| 2 | 52 | 1 | 1 |
+------+------+-----+--------+
On second thought, required skill level and offered skill levels might be selected, and CASE WHEN used to extract a match (1 if offered >= required, offered/required otherwise). The average of the resulting values might then be used as a "skill match" index.
As for the "How is X matched for Y?" question, once you know that candidate CID 1 is good for job JID 50, you can run a LEFT JOIN between the candidates and the jobs. You already know that whatever the jobs asks, the candidate has, but this way you retrieve what unasked-for skills the candidate has, and the other relevant values as well.
A nested query joining the query above and this last one would perhaps tell you everything in one fell swoop, but it would be a costly swoop, I think :-)

Answer 5

If you don't care about cross-dbms support, you can use mysql specific group_concat to reduce the two s_code sets to strings, one per candidate/job and just compare them like this:
select * from
(select cs.cid, group_concat(cs.s_code order by cs.s_code) skills
from candidate_skill cs group by cs.cid) cs,
(select sj.jid, group_concat(sj.s_code order by sj.s_code) skills
from skill_job sj group by sj.jid) sj
where sj.skills = cs.skills
Be aware of that when your tables grown the the final where will be slower and slower and you can't create indexes on the group_concat() created fields.