Question

你好，我需要帮助连接两个我正在使用的查询，以便在一个查询中而不是两个不同的查询中获取结果

第一个查询是：

SELECT p.ID as product_id, p.post_title,
       max( CASE WHEN pm.meta_key = 'product_gender' and p.ID = pm.post_id THEN pm.meta_value END ) as gender, 
       max( CASE WHEN pm.meta_key = '_price' and p.ID = pm.post_id THEN pm.meta_value END ) as price, 
       max( CASE WHEN pm.meta_key = 'product_color' and p.ID = pm.post_id THEN pm.meta_value END ) as color, 
       max( CASE WHEN pm.meta_key = 'product_category' and p.ID = pm.post_id THEN pm.meta_value END ) as category,
       max( CASE WHEN pm.meta_key = 'product_size' and p.ID = pm.post_id THEN pm.meta_value END ) as size
FROM    
  wp_posts p 
  join wp_postmeta pm on p.ID = pm.post_id 
group by 
  p.ID

第二个是：

SELECT parentmeta.post_id as post_id,
 concat((select option_value from wp_options where option_name ='siteurl'  limit 1),'/wp-content/uploads/',childmeta.meta_value)                           as url
 FROM wp_postmeta childmeta 
 INNER JOIN wp_postmeta parentmeta ON (childmeta.post_id=parentmeta.meta_value)
 WHERE parentmeta.meta_key='_thumbnail_id' and childmeta.meta_key = '_wp_attached_file'
 AND parentmeta.post_id = p.ID  //I want to used the same ID as the previous query

现在两个表都给我这个结果

第一个查询：

product_id  post_title   gender    price   color   category   size
83          Puma-Tshirt  Man       19      RED     Tshirt     Medium
86          Nike-Pants   Man       49      BLACK   Pants      Medium

第二个查询：

product_id   url
83           www.img.com/puma.jpg
86           www.img.com/nike.jpg

我想要的是结合查询以得到类似的东西：

product_id  post_title   gender    price   color   category   size   url
83          Puma-Tshirt  Man       19      RED     Tshirt     Medium www.img.com/puma.jpg
86          Nike-Pants   Man       49      BLACK   Pants      Medium www.img.com/nike.jpg

非常感谢所有帮助。

Answer 1

( SELECT p.ID as product_id, p.post_title,
       max( CASE WHEN pm.meta_key = 'product_gender' and p.ID = pm.post_id THEN pm.meta_value END ) as gender, 
       max( CASE WHEN pm.meta_key = '_price' and p.ID = pm.post_id THEN pm.meta_value END ) as price, 
       max( CASE WHEN pm.meta_key = 'product_color' and p.ID = pm.post_id THEN pm.meta_value END ) as color, 
       max( CASE WHEN pm.meta_key = 'product_category' and p.ID = pm.post_id THEN pm.meta_value END ) as category,
       max( CASE WHEN pm.meta_key = 'product_size' and p.ID = pm.post_id THEN pm.meta_value END ) as size
FROM    
  wp_posts p 
  join wp_postmeta pm on p.ID = pm.post_id 
group by 
  p.ID )   
UNION   
(SELECT parentmeta.post_id as post_id,
 concat((select option_value from wp_options where option_name ='siteurl'  limit 1),'/wp-content/uploads/',childmeta.meta_value)                           as url
 FROM wp_postmeta childmeta 
 INNER JOIN wp_postmeta parentmeta ON (childmeta.post_id=parentmeta.meta_value)
 WHERE parentmeta.meta_key='_thumbnail_id' and childmeta.meta_key = '_wp_attached_file'
 AND parentmeta.post_id = p.ID  //I want to used the same ID as the previous query)

这将为您提供所需的结果。
从您的问题来看，就像您只是想将这两个组合结合在一起，而不论它们的联接和所有联接。

Answer 2

您可以加入两个查询

select t1.*, t2.* 
from  (
  SELECT p.ID as product_id, p.post_title,
         max( CASE WHEN pm.meta_key = 'product_gender' and p.ID = pm.post_id THEN pm.meta_value END ) as gender, 
         max( CASE WHEN pm.meta_key = '_price' and p.ID = pm.post_id THEN pm.meta_value END ) as price, 
         max( CASE WHEN pm.meta_key = 'product_color' and p.ID = pm.post_id THEN pm.meta_value END ) as color, 
         max( CASE WHEN pm.meta_key = 'product_category' and p.ID = pm.post_id THEN pm.meta_value END ) as category,
         max( CASE WHEN pm.meta_key = 'product_size' and p.ID = pm.post_id THEN pm.meta_value END ) as size
  FROM wp_posts p 
  join wp_postmeta pm on p.ID = pm.post_id 
  group by  p.ID 
) t1 
inner join  (
  SELECT parentmeta.post_id as post_id,
   concat((select option_value from wp_options where option_name ='siteurl'  limit 1),'/wp-content/uploads/',childmeta.meta_value)                           as url
   FROM wp_postmeta childmeta 
   INNER JOIN wp_postmeta parentmeta ON (childmeta.post_id=parentmeta.meta_value)
   WHERE parentmeta.meta_key='_thumbnail_id' and childmeta.meta_key = '_wp_attached_file'
) t2 on t1.product_id = t2.post_id

Answer 3

查询结构将是-

SELECT 
A.*,B.Url
FROM
(
    --Your First Query
)A
INNER JOIN 
(
    --Your Second Query
)B
ON A.Product_id = B.Product_id

注意：如果查询2中的记录少于查询1，则使用LEFT JOIN。

我需要帮助加入两个查询才能得到一张表

3 个答案: