Inilah salah satu caranya.
Secara teori itu dapat mengatasi hingga 20 tag per Produk (dibatasi oleh ukuran tabel angka) saya tidak repot-repot mencobanya. Di desktop saya, butuh waktu sekitar 30 detik untuk menghasilkan 65.535 hasil untuk satu produk dengan 16 tag. Semoga jumlah tag aktual Anda per produk akan jauh lebih sedikit dari itu!
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
Pengembalian
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
