在WHERE子句中避免多个SELECT

Question

我试图增强在商家网站中找到的SQL错误。

我有两个表：

------ Table "products" ------ 
| id_product | product_name | 
----------------------------- 
|   1 |  T-shirt | 
|   2 |  Trousers | 
|   3 | Sweat-shirt | 
|   4 |  Socks | 

----------- Table "features" ----------- 
| id_product | feature | feature_value | 
---------------------------------------- 
|   1 | Color |   Red | 
|   1 | Size |    M | 
|   1 | Fabric |  Cotton | 
|   2 | Color |   Blue | 
|   2 | Size |    S | 
|   2 | Fabric |  Polyester | 
|   3 | Color |   Red | 
|   3 | Size |    L | 
|   3 | Fabric |   Wool | 
|   4 | Color |   White | 
|   4 | Size |    L | 
|   4 | Fabric |  Cotton | 

我想，具有以下特点检索产品：

• 其中颜色是红色或蓝色
• 其中大小为M
• 其中织物是棉花

我的查询如下：

SELECT p.id_product 
FROM products p 
WHERE p.id_product IN (SELECT f.id_product FROM features f WHERE f.feature_value IN ("Red", "Blue")) 
    AND p.id_product IN (SELECT f.id_product FROM features f WHERE f.feature_value = "M") 
    AND p.id_product IN (SELECT f.id_product FROM features f WHERE f.feature_value = "Cotton") 
GROUP BY p.id_product 

（当然，在现实中，我的表，我的查询是waaaay比这更复杂，我只是专注于问题的一部分）

的多个SELECT在WHERE子句中，如果选择了8个或更多功能，则会导致整个服务器变慢。 有没有办法避免在WHERE子句中做这么多查询？

编辑：举例来说，这里是真正的查询之一：

SELECT p.id_product id_product 
FROM ps_product p 
INNER JOIN ps_category_product cp ON p.id_product = cp.id_product 
INNER JOIN ps_category c ON (c.id_category = cp.id_category AND c.nleft >= 6 AND c.nright <= 7 AND c.active = 1) 
LEFT JOIN ps_stock_available sa ON (sa.id_product = p.id_product AND sa.id_shop = 1) 
INNER JOIN ps_product_shop product_shop ON (product_shop.id_product = p.id_product AND product_shop.id_shop = 1) 
WHERE 1 
AND product_shop.active = 1 
AND product_shop.visibility IN ("both", "catalog") 
AND p.id_manufacturer IN (5,4) 
AND sa.quantity > 0 
AND p.id_product IN (SELECT id_product FROM ps_feature_product fp WHERE fp.id_feature_value = 82) 
AND p.id_product IN (SELECT id_product FROM ps_feature_product fp WHERE fp.id_feature_value = 37248) 
AND p.id_product IN (SELECT id_product FROM ps_feature_product fp WHERE fp.id_feature_value = 181) 
AND p.id_product IN (SELECT id_product FROM ps_feature_product fp WHERE fp.id_feature_value = 37821) 
AND p.id_product IN (SELECT id_product FROM ps_feature_product fp WHERE fp.id_feature_value = 33907) 
AND p.id_product IN (SELECT id_product FROM ps_feature_product fp WHERE fp.id_feature_value = 33902) 
AND p.id_product IN (SELECT id_product FROM ps_feature_product fp WHERE fp.id_feature_value = 70) 
AND p.id_product IN (SELECT id_product FROM ps_feature_product fp WHERE fp.id_feature_value = 76) 
AND p.id_product IN (SELECT id_product FROM ps_feature_product fp WHERE fp.id_feature_value = 291) 
AND p.id_product IN (SELECT id_product FROM ps_feature_product fp WHERE fp.id_feature_value = 75) 
AND p.id_product IN (SELECT id_product FROM ps_feature_product fp WHERE fp.id_feature_value = 44459) 
GROUP BY id_product 

Answer 1

你可以使用JOIN和HAVING条款：

SELECT p.id_product 
FROM products p 
JOIN features f 
    ON p.id_product = f.id_product 
GROUP BY p.id_product 
HAVING COUNT(CASE WHEN f.feature_value IN ('Red', 'Blue') THEN 1 END) > 0 
    AND COUNT(CASE WHEN f.feature_value = 'M' THEN 1 END) > 0 
    AND COUNT(CASE WHEN f.feature_value = 'Cotton' THEN 1 END) > 0; 

LiveDemo

---

甚至更​​短（MySQL的）：

HAVING SUM(f.feature_value IN ('Red', 'Blue')) > 0 
    AND SUM(f.feature_value = 'M') > 0 
    AND SUM(f.feature_value = 'Cotton') > 0; 

Answer 2

试试这个，希望这将有助于。

SELECT p.id_product,count(f.feature_value) 
FROM products p, features f 
where f.id_product=p.id_product 
and f.feature_value in ("Red","Blue") 
or f.feature_value = "M" 
or f.feature_value = "Cotton" 
group by p.id_product 

Answer 3

SELECT 
    P.id_product 
FROM 
    products P 
    INNER JOIN features F 
    ON P.id_product = F.id_product 
WHERE 
    (F.featurure = 'Color' AND F.feature_value IN ('red', 'blue')) 
    OR (F.featurure = 'Size' AND F.feature_value IN ('M')) 
    OR (F.featurure = 'Fabric' AND F.feature_value IN ('Cotton')) 
GROUP BY 
    P.id_product 
HAVING 
    COUNT(DISTINCT F.feature) = 3 

HAVING条件告诉，应至少有3个不同的匹配特征（这是您的搜索字段的计数）。