Question

I have a select query to build and I can’t get it work propery. I was hopeing for a few sudggestions from more advanced MySQL developers.
So my tables are:

 CREATE TABLE IF NOT EXISTS `gv` ( 
`id` int(11) NOT NULL AUTO_INCREMENT,
`option_id` int(11) NOT NULL,
`group_id` int(11) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ;

INSERT INTO `gv` (`id`, `option_id`, `group_id`) VALUES
(1, 1, 1),
(2, 2, 2),
(3, 3, 2),
(4, 4, 3),
(5, 5, 4),
(6, 6, 4); 

CREATE TABLE IF NOT EXISTS `igv` (
 `id` int(11) NOT NULL AUTO_INCREMENT,
 `item_id` int(11) NOT NULL,
 `gv_id` int(11) NOT NULL,
 PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=9 ;

INSERT INTO `igv` (`id`, `item_id`, `gv_id`) VALUES
(1, 1, 1),
(2, 1, 3),
(3, 2, 1),
(4, 2, 2),
(6, 3, 5),
(7, 4, 2),
(8, 2, 6); 

CREATE TABLE IF NOT EXISTS `items` (
  `item_id` int(11) NOT NULL AUTO_INCREMENT,
  PRIMARY KEY (`item_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=6 ;

INSERT INTO `items` (`item_id`) VALUES
(1),
(2),
(3),
(4),
(5); 

Now I will explain what these tables are doing:

1. gv comes from group_values. A group is like a property for an item. Ex: Color.

   • each row holds values for a group. Ex: Red ( option_id ) for group Color ( group_id ).

2. igv comes from item group values. An item is the main entity in my project, like a product.

   • each row holds relations between items and group_values. Ex: An item with Red Color.
   • it’s linked to the gv table by gv_id ( igv.gv_id = gv.id ).

3. items comes from items 🙂 I have simplified for this example to only one column, item_id.

What I need to query:

Well, in my app I have a selected item ( my_item ) ( with 0 or more group values attached ).
I need all the items that have exactly the same group_values defined, or no values specified for my_item groups. In php I can extract all the attached groups as an array with ids, also all the group_values. But I found no way to select what I need to from this database.
I will appreciate very much any input provided into this massive time-consumming select.

Thanks!

A short sketched-up example. On the left we have the item that is matching all the right items ( separated with comma ).

Answer 1

OK, so to start with we need to find all items with a group matching another item:

SELECT igv.item_id, sigv.item_id FROM igv
  INNER JOIN gv ON igv.gv_id = gv.id
  INNER JOIN gv sgv ON gv.group_id = sgv.group_id
  INNER JOIN igv sigv ON sigv.gv_id = sgv.id
  WHERE igv.item_id = items.item_id
    AND sigv.item_id = similar.item_id

We then are only interested when the option does not match:

SELECT igv.item_id, sigv.item_id dissimilar FROM igv
  INNER JOIN gv ON igv.gv_id = gv.id
  INNER JOIN gv sgv ON gv.group_id = sgv.group_id
  INNER JOIN igv sigv ON sigv.gv_id = sgv.id
  WHERE igv.item_id = items.item_id
    AND sigv.item_id = similar.item_id
    AND gv.option_id != sgv.option_id

Then we are only interested in the items that are not selected above. This can be done using either NOT EXISTS or a LEFT JOIN. Here’s how to do it with NOT EXISTS:

SELECT items.item_id, similar.item_id similar_id
  FROM items similar
  INNER JOIN items
  WHERE items.item_id != similar.item_id
    AND NOT EXISTS (
    SELECT igv.item_id, sigv.item_id dissimilar FROM igv
      INNER JOIN gv ON igv.gv_id = gv.id
      INNER JOIN gv sgv ON gv.group_id = sgv.group_id
      INNER JOIN igv sigv ON sigv.gv_id = sgv.id
      WHERE sigv.item_id = similar.item_id
        AND igv.item_id = items.item_id
        AND gv.option_id != sgv.option_id
    )

Result:

ITEM_ID SIMILAR_ID
1       3
1       5
2       4
2       5
3       1
3       4
3       5
4       2
4       3
4       5
5       1
5       2
5       3
5       4