Question

我有2个问题：

select COUNT(o.id_offer) from offers o, product p where
 p.id_product = "+ productID +" and o.id_offer = "+ offerID +" and (b.type = 0 or o.type = "A")

此查询返回某个产品的计数

select p.id_product, p.name, s.id_supplier, s.name from product p, suppliers s where p.id_supplier = s.id_supplier

此查询返回所有产品数据

我想将两个查询合并为一个，如下所示：

select (select COUNT(o.id_offer) from offers o, product p where
 p.id_product = p.id_product and o.id_offer = 13345 and (b.type = 0 or o.type = "A")) count,p.id_product, p.name, s.id_supplier, s.name from product p, suppliers s where p.id_supplier = s.id_supplier

任何人都知道如何在informix中执行此操作？

Answer 1

请使用显式连接表示法，而不是FROM子句表示法中表名的逗号列表。

第一个查询需要进行概括以生成产品ID和匹配计数。

SELECT p.id_product, COUNT(o.id_offer) AS offer_count
  FROM offers o
  JOIN product p ON p.id_product = o.id_product -- Guessed column
 WHERE (p.type = 0 OR o.type = 'A')
 GROUP BY p.id_product

第二个查询可以转换为：

SELECT p.id_product, p.name, s.id_supplier, s.name
  FROM product p
  JOIN suppliers s ON p.id_supplier = s.id_supplier

这两个查询也可以与连接组合使用：

SELECT x.id_product, x.product_name, x.id_supplier, x.supplier_name, y.offer_count
  FROM (SELECT p.id_product, p.name, s.id_supplier, s.name
          FROM product p
          JOIN suppliers s ON p.id_supplier = s.id_supplier
       ) AS x
  JOIN (SELECT p.id_product, COUNT(o.id_offer) AS offer_count
          FROM offers o
          JOIN product p ON p.id_product = o.id_product -- Guessed column
         WHERE (p.type = 0 OR o.type = 'A')
         GROUP BY p.id_product
       ) AS y
    ON x.id_product = y.id_product

Informix：查询合并

1 个答案: