MySQL：查询与另一个表具有给定关系集的所有记录

Question

是的，有点笨拙，但是我发现很难描述我的问题。 我有以下表格：

属性

id | name
1  | color
2  | material

选项

id | property_id | name
1  | 1           | yellow
2  | 1           | blue
3  | 2           | wood
4  | 2           | stone

物质

id | name
1  | orange juice
2  | cheese

关系

id | substance_id | option_id
1  | 2            | 1
2  | 2            | 3
3  | 1            | 1

现在，我有一个选项列表，想知道哪些物质与所有这些选项有关。 （例如，哪些物质是黄色的并且是木头制成的？）通过一个查询可以做到吗？

我正在Rails中尝试这样做。

Answer 1

就在我脑海中，您可以尝试：

SELECT DISTINCT s.name FROM substances s, relation r 
WHERE r.substance_id = s.id 
      AND r.option_id IN ( 1, 3) 

Answer 2

(SELECT s.name FROM substances s, relations r, options o
WHERE r.substance_id = s.id and r.option_id = o.id and o.name='yellow')
INTERSECT
(SELECT s.name FROM substances s, relations r, options o
WHERE r.substance_id = s.id and r.option_id = o.id and o.name='wood')

要么

SELECT s.name FROM substances s
WHERE exists(SELECT * from relations r, options o 
      WHERE r.substance_id = s.id and r.option_id = o.id and o.name='yellow')
AND   exists(SELECT * from relations r, options o 
      WHERE r.substance_id = s.id and r.option_id = o.id and o.name='wood')

仅使用一个带有property_id options表来区分不同类型的选项并不是一个好主意，实际上会使这变得更加困难。 我建议将不同类型的选项分解为不同的表格

MATERIALS(id,name)
COLORS(id,name)

并对每种类型的表使用单独的关系。 在这种情况下，您不需要为每个关系使用单独的表，因为看起来这是一个多（物质）到一个（颜色）关系。

SUBSTANCES(id, name, material_id, color_id)

然后您的查询要简单得多

SELECT s.name FROM substances s, materials m, colors c
WHERE s.color_id = c.id AND m.material_id = m.id
AND m.name = 'wood'
AND c.name = 'yellow'

ActiveRecord应该比前两个更容易处理最后一个查询。

Answer 3

SELECT s.name
FROM   substances AS s
       LEFT JOIN relations AS r1
         ON s.id = r1.substance_id
       INNER JOIN relations AS r2
         ON r1.substance_id = r2.substance_id
            AND r1.option_id < r2.option_id
       LEFT JOIN options AS o1
         ON o1.id = r1.option_id
       LEFT JOIN options AS o2
         ON o2.id = r2.option_id
WHERE  o1.name = 'yellow'
       AND o2.name = 'wood'  

我不是很了解如何优化SQL的性能。 您可能要比较上面（与下面）以及此处发布的其他解决方案的基准。

SELECT s.name
FROM   substances AS s
       LEFT JOIN relations AS r1
         ON s.id = r1.substance_id
       INNER JOIN relations AS r2
         ON r1.substance_id = r2.substance_id
            AND r1.option_id < r2.option_id
       LEFT JOIN options AS o1
         ON o1.id = r1.option_id
     AND o1.name = 'wood'
       LEFT JOIN options AS o2
         ON o2.id = r2.option_id
     AND o2.name = 'yellow'