Question

Well, it’s a very simple question, so I hope you could help me.

I’ve two tables with this structure:

-- -----------------------------------------------------
-- Table `mydb`.`product`
-- -----------------------------------------------------
CREATE  TABLE IF NOT EXISTS `mydb`.`products` (
  `id_product` INT NOT NULL ,
  `name` VARCHAR(45) NOT NULL ,
  PRIMARY KEY (`id_product`) )
ENGINE = InnoDB;

-- -----------------------------------------------------
-- Table `mydb`.`features`
-- -----------------------------------------------------
CREATE  TABLE IF NOT EXISTS `mydb`.`features` (
  `id_feature` INT NOT NULL ,
  `id_product` INT NOT NULL ,
  `description` VARCHAR(45) NOT NULL ,
  PRIMARY KEY (`id_feature`) ,
  INDEX `fk_table2_table1` (`id_product` ASC) ,
  CONSTRAINT `fk_table2_table1`
    FOREIGN KEY (`id_product` )
    REFERENCES `mydb`.`products` (`id_product` )
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;

---

This query returns all the features of the product 5:

select descripcion from features where id_product = 5;

which are:

expensive
strong
tall

Ok, now I want to get all the names of the table ‘products’ that have only all these features.
I tried something like this:

select p.name from products p, features f
where p.id_product = f.id_product and
f.id_feature in 
(select id_feature from features 
where description in
(select description from features
where id_product = 5);

But it also gives me back the products which have less or more features than the three that I’m looking for…

I hope I was clear.
Thanks in advance.

Answer 1

I think what you’re looking for are those products that have all three of those features that don’t have any others. If so, something like this should work:

SELECT P.Id_Product, P.Name
FROM Products P
    JOIN Features F ON P.id_product = F.id_product 
    JOIN Features F2 ON P.id_product = F2.id_product 
WHERE F.id_feature IN (SELECT id_feature FROM Features WHERE id_product = 5)
GROUP BY P.Id_Product, P.Name
HAVING COUNT(DISTINCT F2.id_Feature) = 3

And here is a condensed Fiddle to demonstrate.