php - 一个有点复杂的mysql查询

Question

我目前正在一个新闻数据库网站上工作，我似乎无法创建一个查询来选择 5 篇 HitTest 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 - 包含新闻项的所有评级(id、news_id、rating 等)

现在我的查询应该从表 Rates 中选择平均评级最高和投票最多的 5 个 news_id(因此:我认为按 AVG(Rating) 和 COUNT(*) 排序)。我首先尝试进行查询并立即从新闻表中获取这些 news_ids 的所有信息(使用 WHERE id IN(--查询选择 5 个 HitTest 门的 news_ids--) 子句)但是返回了我的 MySql 版本的错误不能在 WHERE IN 子句子查询中使用 LIMIT。

好吧，我希望你能帮助我解决必须选择这 5 个 news_id 的第一个查询。我现在得到的查询(但不是完全有效)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

或满足于我的脚本的其余部分:

$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

您确定 average_r 和 amt_r 都按降序排列吗？

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