mysql - SQL逻辑: sort result by product of a column and number of colums of a specific type

Question

用户可以按评论分数对我的目录进行排序。以下是通过评论分数抓取目录的 SQL:

   SELECT *, null AS score, '0' AS SortOrder
        FROM products
        WHERE datelive < 0
   UNION
   SELECT p.*, ROUND(SUM(r.score)/COUNT(*)) AS score, '1' AS SortOrder
        FROM products p
        LEFT JOIN reviews r
            ON r.id = p.id
            AND p.datelive > 0
        GROUP BY p.id
        HAVING COUNT(*) >= 5
    UNION
    SELECT p.*, null AS score, '2' AS SortOrder
        FROM products p
        LEFT JOIN reviews r
            ON r.id = p.id
            WHERE p.datelive > 0
        GROUP BY p.id
        HAVING COUNT(*) < 5
    ORDER BY SortOrder ASC, score DESC

它的作用是按顺序获取以下内容:

• datelive 类型的产品 < 0
• datelive 类型的产品 > 0 且 >= 5 条评论
• datelive 类型的产品 > 0 条评论且 < 5 条评论

整个内容也是按照评论分数排序的。我想按原样对中间的要点进行排序，但乘以分数 >= 3.53 的评论百分比。

假设某个产品的原始评论得分为 4.53，但二十条评论中有两条低于 3.53，即 90% 的评论得分 >= 3.53。我想按 (score*mult) 对其进行排序，其中 mult 是评论数 >= 3.53 除以评论总数。在本例中，我排序的分数是 4.53 * 0.9 = 4.08。

Answer 1

将 mult 添加到中间查询，并将 1 作为其他查询中的值。然后在 ORDER BY 子句中使用 score*mult。

此外，MySQL 有一个内置函数用于计算平均值，您不需要除以计数。

   SELECT *, null AS score, '0' AS SortOrder, 1 AS mult
        FROM products
        WHERE datelive < 0
   UNION
   SELECT p.*, ROUND(AVG(r.score)) AS score, '1' AS SortOrder,
            SUM(r.score > 3.53)/count(*) AS mult
        FROM products p
        LEFT JOIN reviews r
            ON r.id = p.id
            AND p.datelive > 0
        GROUP BY p.id
        HAVING COUNT(*) >= 5
    UNION
    SELECT p.*, null AS score, '2' AS SortOrder, 1 AS mult
        FROM products p
        LEFT JOIN reviews r
            ON r.id = p.id
            WHERE p.datelive > 0
        GROUP BY p.id
        HAVING COUNT(*) < 5
    ORDER BY SortOrder ASC, score*mult DESC