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MySQL: select records in which the joined table matches all values

I try to find all employees with several skills. Here are the tables:

CREATE TABLE IF NOT EXISTS `Employee` (
  `ID` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
  `Name` varchar(100) DEFAULT NULL,
  PRIMARY KEY (`ID`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=3 ;
INSERT INTO `Employee` (`ID`, `Name`, `Region_ID`) VALUES (1, 'Fred Flintstone'), (2, 'Barney Rubble');

CREATE TABLE IF NOT EXISTS `Skill` (
  `ID` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
  `Name` varchar(100) DEFAULT NULL,
  PRIMARY KEY (`ID`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=3 ;
INSERT INTO `Skill` (`ID`, `Name`) VALUES (1, 'PHP'), (2, 'JQuery');

CREATE TABLE IF NOT EXISTS `Emp_Skills` (
  `ID` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
  `Emp_ID` bigint(20) unsigned NOT NULL DEFAULT '0',
  `Skill_ID` bigint(20) unsigned NOT NULL DEFAULT '0',
  PRIMARY KEY (`ID`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=5 ;
INSERT INTO `Emp_Skills` (`ID`, `Emp_ID`, `Skill_ID`) VALUES (1, 1, 1), (2, 1, 2), (3, 2, 1);

Here is the query that I still have:

SELECT DISTINCT(em.ID), em.Name 
FROM Employee em 
INNER JOIN Emp_Skills es ON es.Emp_ID = em.ID
WHERE es.Skill_ID IN ('1', '2')

This returns both employees, however I need to find an employee who has both skills (IDs 1 and 2).

Any ideas? Thanks

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2 answers

This will be done:

SELECT EmpId, Name
FROM
(
   SELECT em.ID as EmpId, em.Name, es.ID as SkillID 
   FROM Employee em 
   INNER JOIN Emp_Skills es ON es.Emp_ID = em.ID
   WHERE es.Skill_ID IN ('1', '2')
 ) X
GROUP BY EmpID, Name
HAVING COUNT(DISTINCT SkillID) = 2;

Spell here:

Differences, just in case, the same employee has the skill specified twice.

Thanks for the test data.

+3
source

You can do this with aggregation and a suggestion having:

SELECT em.ID, em.Name 
FROM Employee em INNER JOIN
     Emp_Skills es
     ON es.Emp_ID = em.ID
GROUP BY em.id, em.name
HAVING sum(es.Skill_id = '1') > 0 and
       sum(es.Skill_id = '2') > 0;

having , . .

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