一个有点复杂的mysql查询

Question

我目前正在开发一个新闻数据库网站，但我似乎无法创建查询来选择 5 篇最热门的新闻文章。受此查询影响的数据库的 2 个表是：

• 新闻 - 包含所有新闻项目（id、作者、消息等）
• Rates - 包含新闻项目的所有评级（id、news_id、rating 等）

现在我的查询应该从表 Rates 中选择 5 个 news_ids，具有最高的平均评分和最多的选票（所以：按 AVG(Rating) 和 COUNT(*) 排序）。我首先尝试进行查询，并立即从 News 表中获取这些 news_ids 的所有信息（使用 WHERE id IN(--the query selection the 5 hottest news_ids--) 子句），但这返回了 MySql Version 的错误不能在 WHERE IN 子查询中使用 LIMIT。

好吧，我希望你能帮助我完成第一个必须选择这 5 个 news_id 的查询。我现在得到的查询（但没有完全工作）是：

SELECT news_id FROM 
              (SELECT news_id, AVG(rating) AS average_r, COUNT(*) AS amt_r 
                      FROM rates 
                      GROUP BY news_id 
                      ORDER BY average_r,amt_r 
                      DESC LIMIT 5
               ) AS news_rates

或在我的脚本的其余部分内容：

$get_hot_news_ids = mysql_query("SELECT news_id FROM 
        (SELECT news_id, AVG(rating) AS average_r, COUNT(*) AS amt_r 
         FROM rates 
         GROUP BY news_id 
         ORDER BY average_r,amt_r DESC LIMIT 5) AS news_rates");

    $first = 1;
    while($news_id     = mysql_fetch_assoc($get_hot_news_ids)) {
        if(!$first) {
            $hot_news_ids .= " ,";
        }else{
            $first = 0;
        }
        $hot_news_ids .= $news_id['news_id'];
    }
    //print_r($hot_news_ids);
    $get_hot_news = mysql_query("SELECT * FROM news 
        WHERE id IN($hot_news_ids) 
        ORDER BY FIELD(id, $hot_news_ids)");

Answer 1

你确定average_r和amt_r都是降序排列吗？

SELECT news_id FROM 
              (SELECT news_id, AVG(rating) AS average_r, COUNT(*) AS amt_r 
                      FROM rates 
                      GROUP BY news_id 
                      ORDER BY average_r DESC, amt_r DESC
                      LIMIT 5
               ) AS news_rates

Answer 2

试试这个：

SELECT TOP 5 N.id, N.author, N.message, AVG(R.rating) AS rate, COUNT(R.news_id) AS votes
FROM       news  N
INNER JOIN rates R ON N.id = R.news_id
GROUP BY N.id, N.author, N.message
ORDER BY rate, votes

Answer 3

您可以改用 join，这将允许 limit:

SELECT *
FROM news n JOIN (
    SELECT news_id, AVG(rating) AS average_r, COUNT(*) AS amt_r 
    FROM rates 
    GROUP BY news_id 
    ORDER BY average_r,amt_r DESC 
    LIMIT 5
  ) top5 ON n.news_id = top5.news_id
ORDER BY top5.average_r,top5.amt_r 

注意：您可能希望将查询更改为 ORDER BY average_r DESC, amt_r DESC 以获得最高评级的项目，而不是最低评级。