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我有以下一组 MySQL 查询，用于通过网站跟踪用户进度。有没有简化它们的好方法？

#How many people reached stage 2
SELECT COUNT(DISTINCT a.session_id) as "total"
    FROM formation_page_hits a
    WHERE a.progress = 2
    AND DATE(a.datetime) = "2011-03-23";

#How many people reached stage 4 having reached stage 2
SELECT COUNT(DISTINCT a.session_id) as "total"
    FROM formation_page_hits a, (SELECT f.session_id, f.`datetime`
                        FROM formation_page_hits f
                        WHERE f.progress = 2) as b
    WHERE a.progress = 4
    AND a.session_id = b.session_id
    AND DATE(b.datetime) = "2011-03-23"
    AND DATE(a.datetime) = "2011-03-23";


#How many people reached stage 7, having reached stage 4, having reached stage 2
SELECT COUNT(DISTINCT a.session_id) as "total"
    FROM formation_page_hits a, (SELECT f.session_id, f.`datetime`
                        FROM formation_page_hits f
                        WHERE f.progress = 4) as b, (SELECT f.session_id, f.`datetime`
                        FROM formation_page_hits f
                        WHERE f.progress = 2) as c
    WHERE a.progress = 7
    AND a.session_id = b.session_id
    AND a.session_id = c.session_id
    AND DATE(c.datetime) = "2011-03-23"
    AND DATE(b.datetime) = "2011-03-23"
    AND DATE(a.datetime) = "2011-03-23";

如您所见，我很快重新查询相同的信息，并且有另外 4 或 5 个遵循相同模式的查询 - 有没有更好的方法来构建查询，这意味着我不必继续查询对于“有多少人达到了第 2 阶段”？

编辑：每个页面视图都存储为formation_page_hits 中的一个条目 - 因此每个会话都有完整的页面视图记录

id_formation_page_hits INT PRIMARY_KEY, session_id VARCHAR(100), datetime DATETIME, progress INT
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1 回答 1

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SELECT  COUNT(*)
FROM    (
        SELECT  session_id
        FROM    formation_page_hits
        WHERE   progress IN (2, 4, 7)
                AND datetime >= '2011-03-23'
                AND datetime < '2011-03-24'
        GROUP BY
                session_id
        HAVING  COUNT(DISTINCT progress) = 3
        ) q

为此创建一个复合索引以(session_id, datetime, progress)使其快速工作。

于 2011-03-28T16:04:35.573 回答