我有以下一组 MySQL 查询,用于跟踪用户浏览网站的进度。有什么好的方法可以简化它们吗?
#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,我正在快速地重新查询相同的信息,并且还有另外 4 或 5 个查询遵循相同的模式 - 是否有更好的构造查询的方法,这意味着我没有继续查询“有多少人达到了第 2 阶段”?
编辑:每个页面浏览量都存储为 formation_page_hits 中的一个条目 - 这样每个 session 的页面浏览量都有一个完整的记录
id_formation_page_hits INT PRIMARY_KEY, session_id VARCHAR(100), datetime DATETIME, progress INT
最佳答案
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)
上创建复合索引以使其快速运行。
关于mysql - 组合回收多个自连接的mysql查询,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/5461743/