I'm trying to solve the below problem.
I feel like it is possible, but I can't seem to get it.
Here's the scenario:
Table 1 (Assets)
1 Asset-A
2 Asset-B
3 Asset-C
4 Asset-D
Table 2 (Attributes)
1 Asset-A Red
2 Asset-A Hard
3 Asset-B Red
4 Asset-B Hard
5 Asset-B Heavy
6 Asset-C Blue
7 Asset-C Hard
If I am looking for something having the same attributes as Asset-A, then it should identify Asset-B since Asset-B has all the same attributes as Asset-A (it should discard heavy, since Asset-A didn't specify anything different or the similar). Also, if I wanted the attributes for only Asset-A AND Asset-B that were common, how would I get that?
Seems simple, but I can't nail it...
The actual table I am using, is almost precisely Table2, simply an association of an AssetId, and an AttributeId so:
PK: Id
int: AssetId
int: AttributeId
I only included the idea of the asset table to simplify the question.
SELECT ato.id, ato.value
FROM (
SELECT id
FROM assets a
WHERE NOT EXISTS
(
SELECT NULL
FROM attributes ata
LEFT JOIN
attributes ato
ON ato.id = ata.id
AND ato.value = ata.value
WHERE ata.id = 1
AND ato.id IS NULL
)
) ao
JOIN attributes ato
ON ato.id = ao.id
JOIN attributes ata
ON ata.id = 1
AND ata.value = ato.value
, or in SQL Server 2005
(with sample data to check):
WITH assets AS
(
SELECT 1 AS id, 'A' AS name
UNION ALL
SELECT 2 AS id, 'B' AS name
UNION ALL
SELECT 3 AS id, 'C' AS name
UNION ALL
SELECT 4 AS id, 'D' AS name
),
attributes AS
(
SELECT 1 AS id, 'Red' AS value
UNION ALL
SELECT 1 AS id, 'Hard' AS value
UNION ALL
SELECT 2 AS id, 'Red' AS value
UNION ALL
SELECT 2 AS id, 'Hard' AS value
UNION ALL
SELECT 2 AS id, 'Heavy' AS value
UNION ALL
SELECT 3 AS id, 'Blue' AS value
UNION ALL
SELECT 3 AS id, 'Hard' AS value
)
SELECT ato.id, ato.value
FROM (
SELECT id
FROM assets a
WHERE a.id <> 1
AND NOT EXISTS
(
SELECT ata.value
FROM attributes ata
WHERE ata.id = 1
EXCEPT
SELECT ato.value
FROM attributes ato
WHERE ato.id = a.id
)
) ao
JOIN attributes ato
ON ato.id = ao.id
JOIN attributes ata
ON ata.id = 1
AND ata.value = ato.value
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