Question

I have 4 tables;

• articles
• machines
• features
• required features

the problem is that articles can be produced on machines but only on the machines that have the features that are required by the articles.
a product can require 0 or more features and a machine can have 0 or more features

I want to create a query that shows a complet overview of valid article-machine combinations.
the question is: How can I do that?

Below is an example data set for MySQL. it should result in the following query result:

"car";"virtual machine"
"boat";"virtual machine"
"boat";"lean machine"

here’s the dataset:

CREATE TABLE IF NOT EXISTS `articles` (
  `name` text NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

INSERT INTO `articles` (`name`) VALUES
('car'),
('boat');

CREATE TABLE IF NOT EXISTS `features` (
  `machine` text NOT NULL,
  `feature_name` text NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

INSERT INTO `features` (`machine`, `feature_name`) VALUES
('lean machine', 'punch'),
('virtual machine', 'punch'),
('virtual machine', 'drill');

CREATE TABLE IF NOT EXISTS `machines` (
  `name` text NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

INSERT INTO `machines` (`name`) VALUES
('lean machine'),
('virtual machine');

CREATE TABLE IF NOT EXISTS `required_features` (
  `article` text NOT NULL,
  `feature` text NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

INSERT INTO `required_features` (`article`, `feature`) VALUES
('car', 'drill'),
('boat', 'punch');

Answer 1

With the data you provide, the solution is just a few joins – but I’m assuming that it’s possible for an article to require more than one feature, and for a machine to have one of those features, but not all the features required.

In this case, you would need to also make sure that a machine that has a matching feature of an article also doesn’t lack any other features required by the article. I don’t think a left join would do the job because it would then return the machine for the matching feature, and the null for the non-matching feature… whereas it shouldn’t return the machine at all in this case.

SELECT DISTINCT
    a.name,
    m.name
FROM
    articles a INNER JOIN
    required_features rf ON rf.article = a.name INNER JOIN
    features f ON f.feature_name = rf.feature INNER JOIN
    machines m 
        ON m.name = f.machine
        -- Make sure that there is no feature required that isn't provided by the machine.
        AND NOT EXISTS (
            SELECT 1
            FROM 
                machines m2 INNER JOIN
                features f2 ON f2.machine = m2.name LEFT JOIN
                required_features rf2 ON rf2.feature = f2.feature_name
            WHERE 
                m2.name = m.name AND
                rf2.feature IS NULL
                AND rf2.article = a.name
        )