我有一个简单的Select语句,它从2个MySQL表中返回数据,这很好。我现在需要从第三个相关表中返回一些数据,但不知道如何执行此操作。从其他加入获取相关值的SQL查询
这里是我当前的SQL查询
select
p.ID,
p.post_title,
p.post_name,
max(CASE WHEN pm.meta_key = '_thumbnail_id' THEN pm.meta_value END) as thumbnailID,
CAST(max(CASE WHEN pm.meta_key = '_price' THEN pm.meta_value END) AS UNSIGNED) as Price,
max(CASE WHEN pm.meta_key = '_stock_status' THEN pm.meta_value END) as stockStatus,
max(CASE WHEN pm.meta_key = '_sku' THEN pm.meta_value END) as SKU,
CAST(max(CASE WHEN pm.meta_key = '_sale_price' THEN pm.meta_value END) AS UNSIGNED) as salePrice,
CAST(max(CASE WHEN pm.meta_key = '_regular_price' THEN pm.meta_value END) AS UNSIGNED) as regularPrice,
CAST(max(CASE WHEN pm.meta_key = 'total_sales' THEN pm.meta_value END) AS UNSIGNED) as totalSales
from wp_posts p
join wp_postmeta pm
on p.ID = pm.post_id
where post_type = 'product'
group by p.ID
我现在需要得到来自wp_postmeta表中的值,其中在wp_postmeta表_thumbnail_id的价值在wp_postmeta表和该POST_ID值的值相匹配meta_key值= _wp_attached_file。
下面是来自wp_postmeta表中的一些记录:
的第一条记录有_thumbnail_id = 100.我现在需要从wp_postmeta表中获取记录的值(同一个表),其中POST_ID = 100和meta_key =如下所示_wp_attached_file:
我知道thumbnail_id作为在我的当前查询返回THUMBNAILID - 只是不知道如何再次将它加入wp_postmeta表。
为什么你需要再次加入..任何想法? –
加入您的查询和wp_postmeta。 'select t。*,att.meta_value from(_your query_)as t left join wp_postmeta as att on att.post_id = t.thumbnail_id and att.meta_key ='_ wp_attached_file'' – Serg