Question

我有三张桌子：

产品

PRODUCT_ID|NAME
     1    |Elaborado  
     2    |Adidas

PRODUCT_ATTRIBUTE

 PRODUCT_ATTRIBUTE_ID|PRODUCT_ID|ATTRIBUTE_ID|VALUE
         1           |1         |1           |Solid
         2           |1         |2           |Casual
         3           |2         |1           |Solid

ATTRIBUTE

ATTRIBUTE_ID|DESCRIPTION
         1  |Pattern
         2  |Occasion

所有属性都在ATTRIBUTE表中。产品和属性之间的链接在PRODUCT_ATTRIBUTE中。现在我需要根据属性过滤产品：

对于前：我需要的产品的图案是“坚实的”，而且场合是'休闲'：

结果应如下所示：

PRODUCT_ID
----------
1

我正在使用Mysql数据库。我准备了一个SQLFiddle：http://sqlfiddle.com/#!9/aa2be/3

请帮忙。

谢谢，窗扇。

Answer 1

看看这个

select group_concat(product.name),attribute.DESCRIPTION,product_attribute.value  from product
join product_attribute
on (product_attribute.product_id=product.product_id)
join attribute
on (product_attribute.attribute_id=attribute.attribute_id)
where (attribute.attribute_id = 1 and product_attribute.value = 'Solid')
or 
(attribute.attribute_id = 2 and product_attribute.value = 'Casual')
group by attribute.attribute_id, product_attribute.value

Answer 2

将列存储为行的概念称为Entity-Attribute-Value (EAV)，它会产生复杂的查询:)例如，您可以找到＆＃34; pattern = solid和occasion = casual＆＃34;像：

select  p.product_id
from    product p
join    product_attribute pa1
on      p.product_id = pa1.product_id
join    attribute a1
on      pa1.attribute_id = a1.attribute_id
join    product_attribute pa2
on      p.product_id = pa2.product_id
join    attribute a2
on      pa2.attribute_id = a2.attribute_id
where   a1.description = 'Pattern'
        and pa1.value = 'Solid'
        and a2.description = 'Occasion'
        and pa2.value = 'Casual'

Example at SQL Fiddle.

另一种方法是创建一个透视视图，例如：

create view vw_product_attribute as
select  p.product_id
,       pa_pattern.value as pattern
,       pa_occasion.value as occasion
from    product p
join    product_attribute pa_pattern
on      p.product_id = pa_pattern.product_id
join    attribute a_pattern
on      pa_pattern.attribute_id = a_pattern.attribute_id
join    product_attribute pa_occasion
on      p.product_id = pa_occasion.product_id
join    attribute a_occasion
on      pa_occasion.attribute_id = a_occasion.attribute_id
where   a_pattern.description = 'Pattern'
        and a_occasion.description = 'Occasion'

然后您可以查询：

select  product_id
from    vw_product_attribute
where   pattern = 'Solid'
        and occasion = 'Casual'

Example of the alternate approach at SQL Fiddle.

Answer 3

这对你有用吗？您需要在设置了patternSolidProduct变量的子查询和SELECT语句上设置描述和值。

您还应该在attribute。attribute_id，product。product_id和product_attribute上添加PK索引。product_attribute_id

我将索引添加到product_attribute。value，长度为10，attribute。description，长度为10。

/*Start join method */
/* Get product id's matching pattern = solid */
SET @patternSolidProduct := (SELECT
  GROUP_CONCAT(`pa`.`product_id`)
FROM
  `product_attribute` AS `pa`
LEFT JOIN
  `product` as `p`
ON
  `pa`.`product_id` = `p`.`product_id`
LEFT JOIN
  `attribute` AS `a`
ON
  `a`.`attribute_id` = `pa`.`attribute_id`
WHERE
  `a`.`description` = 'Pattern'
    AND `pa`.`value` = 'Solid'
);
/* End product id's matching pattern = solid */

SELECT
  `pa`.`product_id`
FROM
  `product_attribute` AS `pa`
LEFT JOIN
  `product` as `p`
ON
  `pa`.`product_id` = `p`.`product_id`
LEFT JOIN
  `attribute` AS `a`
ON
  `a`.`attribute_id` = `pa`.`attribute_id`
WHERE
  `a`.`description` = 'Occasion'
  AND `pa`.`value` = 'Casual'
  AND `p`.`product_id` IN (@patternSolidProduct)
/*End join method */

上述解释：

# id, select_type, table, type, possible_keys, key, key_len, ref, rows, Extra
'1', 'SIMPLE', 'p', 'const', 'PRIMARY', 'PRIMARY', '8', 'const', '1', 'Using index'
'1', 'SIMPLE', 'pa', 'ref', 'value', 'value', '32', 'const', '1', 'Using where'
'1', 'SIMPLE', 'a', 'eq_ref', 'PRIMARY,description', 'PRIMARY', '8', 'sqlfiddle.pa.attribute_id', '1', 'Using where'

SQL过滤复杂查询

3 个答案: