Her er én måde.
I teorien kan den klare op til 20 tags pr. produkt (begrænset af størrelsen på taltabellen). Det gad jeg dog ikke prøve. På mit skrivebord tog det omkring 30 sekunder at få de 65.535 resultater for et enkelt produkt med 16 tags. Forhåbentlig vil dit faktiske antal tags pr. produkt være meget mindre end det!
IF OBJECT_ID('tempdb..#Nums') IS NULL
BEGIN
CREATE TABLE #Nums
(
i int primary key
)
;WITH
L0 AS (SELECT 1 AS c UNION ALL SELECT 1),
L1 AS (SELECT 1 AS c FROM L0 A CROSS JOIN L0 B),
L2 AS (SELECT 1 AS c FROM L1 A CROSS JOIN L1 B),
L3 AS (SELECT 1 AS c FROM L2 A CROSS JOIN L2 B),
L4 AS (SELECT 1 AS c FROM L3 A CROSS JOIN L3 B),
L5 AS (SELECT 1 AS c FROM L4 A CROSS JOIN L4 B),
Nums AS (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT 0)) AS i FROM L5)
INSERT INTO #Nums
SELECT TOP 1048576 i FROM Nums;
END
;with ProductTags As
(
SELECT 1 ProductId,'Leather' AS Tag UNION ALL
SELECT 1, 'Watch' UNION ALL
SELECT 2, 'Red' UNION ALL
SELECT 2, 'Necklace' UNION ALL
SELECT 2, 'Pearl'
), NumberedTags AS
(
SELECT
ProductId,Tag,
ROW_NUMBER() OVER (PARTITION BY ProductId ORDER BY Tag) rn,
COUNT(*) OVER (PARTITION BY ProductId) cn
FROM ProductTags
),
GroupedTags As
(
SELECT ProductId,Tag,i
FROM NumberedTags
JOIN #Nums on
#Nums.i < POWER ( 2 ,cn)
and #Nums.i & POWER ( 2 ,rn-1) > 0
)
SELECT ProductId,
STUFF((SELECT CAST(', ' + Tag AS VARCHAR(MAX))
FROM GroupedTags g2
WHERE g1.ProductId = g2.ProductId and g1.i = g2.i
ORDER BY Tag
FOR XML PATH ('')),1,1,'') AS Tags
FROM GroupedTags g1
GROUP BY ProductId, i
ORDER BY ProductId, i
Returnerer
ProductId Tags
----------- ------------------------------
1 Leather
1 Watch
1 Leather, Watch
2 Necklace
2 Pearl
2 Necklace, Pearl
2 Red
2 Necklace, Red
2 Pearl, Red
2 Necklace, Pearl, Red