我有三张桌子。调色板,颜色和关系表palette_color。就像这个样本:
http://sqlfiddle.com/#!6/fe832/2
我想计算关系表中具有相同颜色的调色板。正如您在示例中看到的那样,我已经在做了。但我相信我的方法效率不高。运行需要将近2秒钟。
我正在使用SQL Server。
这是我计算行数的地方:
(
SELECT count(DISTINCT palette_id) as total FROM palette_color COLOR
WHERE NOT EXISTS
(( (SELECT color_id FROM palette_color WHERE palette_id = PALETTE.id) EXCEPT (SELECT color_id FROM palette_color WHERE palette_id = COLOR.palette_id) )
UNION ALL
( (SELECT color_id FROM palette_color WHERE palette_id = COLOR.palette_id) EXCEPT (SELECT color_id FROM palette_color WHERE palette_id = PALETTE.id) ))
) as total
在where子句中,我确保只有第一个调色板出现在结果
上WHERE id =
(
SELECT MIN(palette_id) FROM palette_color COLOR
WHERE NOT EXISTS
(( (SELECT color_id FROM palette_color WHERE palette_id = PALETTE.id) EXCEPT (SELECT color_id FROM palette_color WHERE palette_id = COLOR.palette_id) )
UNION ALL
( (SELECT color_id FROM palette_color WHERE palette_id = COLOR.palette_id) EXCEPT (SELECT color_id FROM palette_color WHERE palette_id = PALETTE.id) ))
)
答案 0 :(得分:4)
在这里,我使用color_id
palete_id
中所有FOR XML PATH
的字符串列表
然后分组并计算每组颜色。
SQL FIDDLE DEMO(12ms)
with cList as (
SELECT p.id palette_id,
STUFF(( SELECT ',' + CAST(pc.color_id as varchar(10) )
FROM palette_color pc
WHERE pc.palette_id = p.id
ORDER BY pc.color_id
FOR
XML PATH('')
), 1, 1, '') AS ColorList
FROM palette p
)
select min(palette_id) palette_id, ColorList, count(*) Total
from cList
group by ColorList
答案 1 :(得分:2)
您可以使用FOR XML PATH执行此操作:
查询1 :
SELECT MIN(palette_id), count(*), colors
FROM (
SELECT id as palette_id,
colors = STUFF((
SELECT ',' + convert(nvarchar(20), pc.color_id)
FROM palette_color pc
WHERE pc.palette_id = palette.id
ORDER BY pc.color_id
FOR XML PATH(''), TYPE).value('.', 'NVARCHAR(MAX)'), 1, 1, '')
FROM palette) a
GROUP BY colors
<强> Results 强>:
| | | colors |
|----|----|-----------|
| 1 | 15 | 1,2 |
| 6 | 60 | 1,2,3 |
| 26 | 6 | 1,2,3,4 |
| 46 | 42 | 1,2,3,4,5 |
| 28 | 18 | 1,3,4 |
| 34 | 36 | 1,3,4,5 |
作为奖励,此解决方案为您提供调色板中使用的实际颜色