Question

表单和mysql有困难。 3个表，1个表值的总和。表单提供了搜索的价值，但它不适用于“WHERE＆gt; ='$ search_total_rating”在错误的地方，我在这里做了一些非常错误。

        $result = mysql_query("SELECT coffeeshops.*, services.*, ratings.*, sum(temp.total) as          final_total FROM coffeeshops inner join services on coffeeshops.shop_id=services.shop_id
inner join ratings on coffeeshops.shop_id=ratings.shop_id
    inner join (select SUM(comfort + service + ambience + friendliness + spacious)/(5) / COUNT(shop_id) AS total, shop_id FROM ratings GROUP BY shop_id) as temp on coffeeshops.shop_id=temp.shop_id WHERE >= '$search_total_rating'");

我不完全理解这一点，但我想要做的是总评级总和> =选择评级。我试图访问final_total，这不是我的数据库中的实际列，这就是为什么使用SUM来获得每个商店的总评级。希望这是代码的一个小小的改组。感谢

Answer 1

您应该使用而不是

SELECT coffeeshops.*, services.*, ratings.*, sum(temp.total) as final_total 
FROM coffeeshops inner join services on coffeeshops.shop_id=services.shop_id
inner join ratings on coffeeshops.shop_id=ratings.shop_id
inner join (
      select SUM(comfort + service + ambience + friendliness + spacious)/5/    COUNT(shop_id) AS total, shop_id 
FROM ratings GROUP BY shop_id) 
as temp on  coffeeshops.shop_id=temp.shop_id 
 having final_total >= '$search_total_rating'

Answer 2

您已在子查询中计算了总计。无需第二个SUM()。

SELECT coffeeshops.*
     , services.*
     , ratings.*
     , temp.total as final_total 
FROM coffeeshops
  inner join services
    on coffeeshops.shop_id = services.shop_id
  inner join ratings 
    on coffeeshops.shop_id = ratings.shop_id
  inner join
    ( select SUM(comfort + service + ambience + friendliness + spacious) / 5
             / COUNT(shop_id) AS total
           , shop_id 
      FROM ratings 
      GROUP BY shop_id
    ) as temp 
    on coffeeshops.shop_id = temp.shop_id 
WHERE temp.total >= '$search_total_rating'

您还可以在子查询中使用HAVING：

SELECT coffeeshops.*
     , services.*
     , ratings.*
     , temp.total as final_total 
FROM coffeeshops
  inner join services
    on coffeeshops.shop_id = services.shop_id
  inner join ratings 
    on coffeeshops.shop_id = ratings.shop_id
  inner join
    ( select SUM(comfort + service + ambience + friendliness + spacious) / 5
             / COUNT(shop_id) AS total
           , shop_id 
      FROM ratings 
      GROUP BY shop_id
      HAVINGE total >= '$search_total_rating'
    ) as temp 
    on coffeeshops.shop_id = temp.shop_id

Mysql查询 - JOINS和WHERE子句，初学者在这里:)

2 个答案: