Question

我需要从product_attributes表中获取所有产品我有几对属性id和过滤器文本值。

由于文本存储为由;分隔的数组，因此更加困难。

       TABLE product_attribute

        Fields product_id, attribute_id, text

        SELECT  pa.product_id  FROM  product_attribute as pa  

WHERE   (
 (pa.attribute_id = '15' AND pa.text LIKE '%Male%' ) 
  AND 
(pa.attribute_id = '12'  AND  pa.text  LIKE  '%Cream%'  )
 )

显然这不起作用，因为attribute_id不能同时为12和15。

我不能在这里使用OR，因为它会退回所有产品（所有男性）+（所有奶油）

我只需要交叉（男性奶油）

在这里得到答案。 Alternative to Intersect in MySQL

我的变体：

SELECT  paz.product_id 
FROM  (
( SELECT product_id FROM oc_product_attribute WHERE attribute_id = '15' AND text LIKE '%male%' )   
UNION ALL
( SELECT product_id FROM oc_product_attribute WHERE attribute_id = '12'  AND text  LIKE  '%creme%' )  
)  paz GROUP BY paz.product_id  HAVING COUNT(*)=2

戈登林诺夫变种^

SELECT pa.product_id 
FROM product_attribute pa 
GROUP BY pa.product_id
HAVING SUM(pa.attribute_id = '15' AND pa.text LIKE '%Male%') > 0 AND
       SUM(pa.attribute_id = '12' AND pa.text LIKE '%Cream%');

Answer 1

我经常使用having和SELECT pa.product_id FROM product_attribute pa GROUP BY pa.product_id HAVING SUM(pa.attribute_id = '15' AND pa.text LIKE '%Male%') > 0 AND SUM(pa.attribute_id = '12' AND pa.text LIKE '%Cream%') > 0;：

来处理这些问题

HAVING

exec dbo.[SP_ViewAdvanceSearchDetails] NULL,NULL,NULL,'AP','Saudi','1','2','2','Looking',NULL子句中的每个条件都验证至少有一行符合特定条件。

Answer 2

如果您只有2个要比较的属性，请使用自我加入product_attribute

SELECT 
  pa.product_id 
FROM
  product_attribute AS pa 
  JOIN product_attribute AS pa1 USING (product_id) 
WHERE pa.attribute_id = '15' 
  AND pa.text LIKE '%Male%' 
  AND pa1.attribute_id = '12' 
  AND pa1.text LIKE '%Cream%'

另一种方法是使用COUNT/GROUP BY /HAVING

SELECT 
  pa.product_id 
FROM
  product_attribute AS pa 
WHERE pa.attribute_id IN ('15', '12') 
  AND (
    pa.text LIKE '%Male%' 
    OR pa.text LIKE '%Cream%'
  ) 
GROUP BY pa.product_id 
HAVING COUNT(DISTINCT pa.attribute_id) = 2

Answer 3

SELECT  pa.product_id  FROM  product_attribute as pa  
WHERE   
(pa.attribute_id = '15' AND pa.text LIKE '%Male%' ) 
union
SELECT  pa.product_id  FROM  product_attribute as pa  
WHERE   
(pa.attribute_id = '12' AND pa.text LIKE '%Male%' )

与工会结合......

你也可以一起做所有事情并区分结果

select a.* from 
(select * from product_attribute where pa.attribute_id = '15' or 
pa.attribute_id = '12')  as a 
where  a.text LIKE '%Male%' or a.text = '%Cream%'

mysql按值对

3 个答案: