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このようなクエリを実行しているときに、mysqlサーバーから高速応答を取得するのに少し問題があります。

select distinct(products.id) as product_id,products.title as product_name, (select count(id) from stock where stock.available='0' and stock.product_id=products.id) as none,
(select count(id) from stock where stock.available='1' and stock.product_id=products.id) as available,
(select count(id) from stock where stock.available='2' and stock.product_id=products.id) as staged,
(select count(id) from stock where stock.available='3' and stock.product_id=products.id) as departed,
(select count(id) from stock where stock.available='4' and stock.product_id=products.id) as delivered
from products,stock where products.id=stock.product_id;

はるかに高速な応答を提供する他のクエリ方法があるのだろうか。ありがとう:-)

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1 に答える 1

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このようなもの：

SELECT
  P.id as product_id,
  P.title as product_name,
  SUM(CASE WHEN S.available = 0 THEN 1 ELSE 0 END) as none,
  SUM(CASE WHEN S.available = 1 THEN 1 ELSE 0 END) as available,
  SUM(CASE WHEN S.available = 2 THEN 1 ELSE 0 END) as staged,
  SUM(CASE WHEN S.available = 3 THEN 1 ELSE 0 END) as departed,
  SUM(CASE WHEN S.available = 4 THEN 1 ELSE 0 END) as delivered 
FROM products P
      JOIN stock S
          ON P.id = S.product_id
    GROUP BY P.id,
             P.title
于 2012-11-22T15:35:25.127 に答える