1

我有这个 mysql 代码，可以通过它们在数据库中出现的次数来检索每个类别中最流行的描述 -

它抓取了列表，但它会像这样回显数据

"Adidas was the most popular in category 5 with 1 occurrences"

我的这个兴趣（爱好）的表结构是

 # interests
 CREATE TABLE `interests` (
 `interestID` int(11) NOT NULL AUTO_INCREMENT,
 `name` varchar(100) NOT NULL,
 `categoryID` int(11) NOT NULL,
 `sessionID` int(11) NOT NULL,
 `interest_desc` varchar(30) NOT NULL,
 `timestamp` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
 PRIMARY KEY (`interestID`)
) ENGINE=MyISAM AUTO_INCREMENT=9 DEFAULT CHARSET=utf8   


# categories
CREATE TABLE `categories` (
 `categoryID` int(11) NOT NULL AUTO_INCREMENT,
 `category_desc` varchar(100) NOT NULL,
 PRIMARY KEY (`categoryID`)
) ENGINE=MyISAM AUTO_INCREMENT=2 DEFAULT CHARSET=utf8

这是我的 SQL 命令：

SELECT 
   interest_desc, 
   categoryID, 
   MAX(num_in_cat) AS num_in_cat,
   category_name
FROM
(
   SELECT interest_desc, categoryID, COUNT(categoryID) AS num_in_cat
   FROM interests
   GROUP BY interest_desc, categoryID
) subsel
JOIN categories ON subsel.categoryID = categories.ID 
GROUP BY interest_desc, categoryID, category_name

我收到此错误：

不是唯一的表/别名：“类别”

4

2 回答 2

1

对您的categories表进行内部连接（您提供正确的表名和列名）是完成此操作所需的全部内容：

SELECT 
  interest_desc, 
  categoryID, 
  MAX(num_in_cat) AS num_in_cat,
  category_name
FROM
(
   SELECT interest_desc, categoryID, COUNT(categoryID) AS num_in_cat
   FROM interests
   GROUP BY interest_desc, categoryID
) subsel
JOIN categories ON subsel.categoryID = categories.ID 
GROUP BY interest_desc, categoryID, category_name
于 2012-05-01T01:32:26.073 回答
0

试试这些：

  1. 如果可行，则更佳，加入

    SELECT s.interest_desc, c.name, MAX(s.num_in_cat) AS num_in_cat FROM ( SELECT interest_desc, categoryID, COUNT(categoryID) AS num_in_cat FROM interests GROUP BY interest_desc, categoryID ) s INNER JOIN categories c ON c.ID = s.categoryID GROUP BY s.interest_desc, s.categoryID

    或者

  2. 添加另一个选择作为列的最佳选择

    SELECT s.interest_desc, (SELECT name from categories c WHERE c.ID = s.categoryID), MAX(s.num_in_cat) AS num_in_cat FROM ( SELECT interest_desc, categoryID, COUNT(categoryID) AS num_in_cat FROM interests GROUP BY interest_desc, categoryID ) s GROUP BY s.interest_desc, s.categoryID

于 2012-05-01T01:32:11.860 回答