结合回收多个自加入的mysql查询

Question

我有以下一组MySQL查询，用于跟踪用户通过网站的进度。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"; 

正如你所看到的，我很快重新查询相同的信息，也有遵循相同的模式附加4个或5个查询 - 在那里建造这意味着我查询一个更好的办法不必一直询问“有多少人到达第二阶段”？

编辑：每个页面视图存储为formation_page_hits的条目 - 因此存在的页面浏览量为每个会话

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 

上(session_id, datetime, progress)为此创建一个综合指数的完整记录快速工作。