MySQL查詢更新表A基於表由乙

Question

我想根據表B.取得

我想[可見：YES]顯示邏輯測試編寫一個MySQL查詢，更新表A邏輯測試只折扣> 40％的產品

折扣邏輯測試：[100 /（old_price/price）]> 40。查詢是對phpMyAdmin的（WordPress的）使用

Table A (product status) 

product_id visible 
1   yes 
2   no 
3   yes 
4   no 

Table B (product details) 

product_id meta_key meta_value 
2   price  550 
2   old_price 600 
1   price  200 
1   old_price 400 
4   price  300 
4   old_price 350 
3   price  100 
3   old_price 300 

Answer 1

這就是答案Dhruv直升機Saxena先生先生親切地送我和完美的作品：

UPDATE product_status ps 
INNER JOIN product_details pd1 
    ON ps.product_id = pd1.product_id 
    AND pd1.meta_key = 'price' 
INNER JOIN product_details pd2 
    ON ps.product_id = pd2.product_id 
    AND pd2.meta_key = 'old_price' 

SET ps.visible = 
      CASE 
       WHEN (100/(pd2.meta_value/pd1.meta_value)) > 40.00 
        THEN 'yes' 
       ELSE 
        'no' 
      END; 

他甚至給我發了一個演示實驗：http://rextester.com/OWRQZE95139

Answer 2

這可能給你想要的東西：

select product_status.product_id, 
case when (1.00*c.meta_value)/(1.00*b.meta_value) < 0.6 then 1 else 0 end as visible 
from product_status 
inner join 
(select * 
from product_details 
where meta_key = 'old_price') b 
on product_status.product_id = b.product_id 
inner join 
(select * 
from product_details 
where meta_key = 'new_price') c 
on product_status.product_id = c.product_id 

如果是的話，用正確的UPDATE語句SELECT語句。

Answer 3

update product_status 
set visible = 'yes' 
where product_id in (select product_id, (100/(max(old_price)/max(price))) as discount 
         from (select product_id, meta_value as old_price, null as price 
           from product_details 
           where meta_key = 'old_price' 
           union 
           select product_id, null, meta_value 
           from product_details 
           where meta_key = 'price') as checkit 
         where (100/(max(old_price)/max(price)) > 40 
         group by product_id));