Question

我有以下一组MySQL查询，用于跟踪用户在网站上的进度。有没有一种简化它们的好方法？

#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";

正如您所看到的，我很快就会重新查询相同的信息，并且还有其他4或5个查询遵循相同的模式 - 是否有更好的方法来构建查询，这意味着我没有继续询问“有多少人到达第二阶段”？

编辑：每个页面视图都存储为formation_page_hits中的条目 - 以便每个会话都有完整的页面查看记录

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

Answer 1

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

在(session_id, datetime, progress)上创建一个复合索引，以便快速工作。

结合回收多个自联接的mysql查询

1 个答案: