Mysql查询-连接和WHERE子句，初学者：）

在表单和mysql方面有困难。3个表，1个表值的总和。表单提供了要搜索的值，但它不适用于“WHERE>=”$search\u total\u rating”的位置错误，我在这里做了一些非常错误的事情

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

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我不完全理解这一点，但我试图做的是总评级总和>=选定评级。我试图访问final_total，它不是我数据库中的实际列，这就是为什么要使用SUM来获得每个商店的总评级。希望这只是对代码的一个小改动。谢谢

您应该使用have而不是where

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'

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您已经在子查询中计算了总计。不需要第二次

SUM（）

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您还可以在子查询中使用

have

：

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 

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部分但不是全部，下面的帖子修复了它，我想：）修复：）非常感谢，所以它使用了2个SUM。非常感谢，我又少了几天闲逛的时间：）

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