【发布时间】：2010-12-09 11:09:22
【问题描述】：
我有一个表格，每次加载网页时都会记录用户 ID、课程、会话 ID 和请求日期。 我想计算给定课程 ID 的每个用户 ID 的持续时间。 由于时间跨度重叠，这样做是有问题的。
此处提供的数据应导致课程 1 的每位用户持续 10 分钟。 我似乎无法做到这一点。
CREATE TABLE PageLogSample (
id INT NOT NULL PRIMARY KEY IDENTITY
, userid INT
, courseid INT
, sessionid INT
, requestdate DATETIME
);
TRUNCATE TABLE PageLogSample;
INSERT INTO PageLogSample (userid, courseid, sessionid, requestdate)
-- [0, 10] = 10 minutes
SELECT 1, 1, 1, '00:00:00'
UNION ALL SELECT 1, 1, 1, '00:10:00'
-- [0, 12] - [3, 5] = 10 minutes
-- or ... [0, 3] + [5, 12] = 10 minutes
UNION ALL SELECT 2, 1, 2, '00:00:00'
UNION ALL SELECT 2, 2, 2, '00:03:00'
UNION ALL SELECT 2, 2, 2, '00:05:00'
UNION ALL SELECT 2, 1, 2, '00:12:00'
-- [0, 12] - [3, 5] = 10 minutes
-- or ... [0, 3] + [5, 12] = 10 minutes
UNION ALL SELECT 3, 1, 3, '00:00:00'
UNION ALL SELECT 3, 2, 3, '00:03:00'
UNION ALL SELECT 3, 2, 3, '00:05:00'
UNION ALL SELECT 3, 1, 3, '00:12:00'
UNION ALL SELECT 3, 2, 3, '00:15:00'
-- [1, 13] - [3, 5] = 10 minutes
-- or ... [1, 3] + [5, 13] = 10 minutes
UNION ALL SELECT 4, 2, 4, '00:00:00'
UNION ALL SELECT 4, 1, 4, '00:01:00'
UNION ALL SELECT 4, 2, 4, '00:03:00'
UNION ALL SELECT 4, 2, 4, '00:05:00'
UNION ALL SELECT 4, 1, 4, '00:13:00'
UNION ALL SELECT 4, 2, 4, '00:15:00'
-- [0, 5] + [10, 15] = 10 minutes
UNION ALL SELECT 5, 1, 5, '00:00:00'
UNION ALL SELECT 5, 1, 5, '00:05:00'
UNION ALL SELECT 5, 1, 6, '00:10:00'
UNION ALL SELECT 5, 1, 6, '00:15:00'
-- [0, 10] = 10 minutes (ignoring everything inbetween)
UNION ALL SELECT 6, 1, 7, '00:00:00'
UNION ALL SELECT 6, 1, 7, '00:03:00'
UNION ALL SELECT 6, 1, 7, '00:05:00'
UNION ALL SELECT 6, 1, 7, '00:07:00'
UNION ALL SELECT 6, 1, 7, '00:10:00'
-- [0, 11] - [5, 6] = 10 minutes
-- or ... [0, 3] + [7, 11] = 6 minutes (good)
-- or ... [0, 5] + [7, 11] = 9 minutes (better)
UNION ALL SELECT 7, 1, 8, '00:00:00'
UNION ALL SELECT 7, 1, 8, '00:03:00'
UNION ALL SELECT 7, 2, 8, '00:05:00'
UNION ALL SELECT 7, 2, 8, '00:06:00'
UNION ALL SELECT 7, 1, 8, '00:07:00'
UNION ALL SELECT 7, 1, 8, '00:11:00'
-- [0, 1] + [2, 4] + [5, 7] + [8, 13] = 10
UNION ALL SELECT 8, 1, 9, '00:00:00'
UNION ALL SELECT 8, 2, 9, '00:01:00'
UNION ALL SELECT 8, 1, 9, '00:02:00'
UNION ALL SELECT 8, 1, 9, '00:03:00'
UNION ALL SELECT 8, 2, 9, '00:04:00'
UNION ALL SELECT 8, 1, 9, '00:05:00'
UNION ALL SELECT 8, 1, 9, '00:06:00'
UNION ALL SELECT 8, 2, 9, '00:07:00'
UNION ALL SELECT 8, 1, 9, '00:08:00'
UNION ALL SELECT 8, 1, 9, '00:13:00'
;
首先尝试天真的方法。这会导致会话的重叠部分出错。
DECLARE @courseid INT;
SET @courseid = 1;
SELECT subquery.userid
, COUNT(DISTINCT subquery.sessionid) AS sessioncount
, SUM(subquery.duration) AS duration
, CASE SUM(subquery.duration)
WHEN 10 THEN 'ok'
ELSE 'ERROR'
END
FROM (
SELECT userid
, sessionid
, DATEDIFF(MINUTE, MIN(requestdate), MAX(requestdate)) AS duration
FROM PageLogSample
WHERE courseid = @courseid
GROUP BY userid
, sessionid
) subquery
GROUP BY subquery.userid
ORDER BY subquery.userid;
-- userid sessioncount duration
-- 1 1 10 ok
-- 2 1 12 ERROR
-- 3 1 12 ERROR
-- 4 1 12 ERROR
-- 5 2 10 ok
第二次尝试。避免重叠。这仅部分有效。
DECLARE @courseid INT;
SET @courseid = 1;
WITH cte (userid, courseid, sessionid, start, finish, duration)
AS (
SELECT userid
, courseid
, sessionid
, MIN(requestdate)
, MAX(requestdate)
, DATEDIFF(MINUTE, MIN(requestdate), MAX(requestdate))
FROM PageLogSample
GROUP BY userid
, courseid
, sessionid
)
SELECT naive.userid
, naive.sessioncount
, naive.duration AS naiveduration
, correction.duration AS correctionduration
, naive.duration - ISNULL(correction.duration, 0) AS duration
, CASE naive.duration - ISNULL(correction.duration, 0)
WHEN 10 THEN 'ok'
ELSE 'ERROR'
END
FROM (
SELECT cte.userid
, COUNT(DISTINCT cte.sessionid) AS sessioncount
, SUM(cte.duration) AS duration
FROM cte
WHERE cte.courseid = @courseid
GROUP BY cte.userid
) naive
LEFT JOIN (
SELECT errors.userid
, SUM(errors.duration) AS duration
FROM cte errors
WHERE errors.courseid <> @courseid
AND EXISTS (
SELECT *
FROM cte
WHERE cte.start <= errors.start
AND cte.finish >= errors.finish
AND cte.courseid = @courseid
)
GROUP BY errors.userid
) correction
ON naive.userid = correction.userid
;
-- userid sessioncount naiveduration correctionduration duration
-- 1 1 10 NULL 10 ok
-- 2 1 12 2 10 ok
-- 3 1 12 NULL 12 ERROR
-- 4 1 12 NULL 12 ERROR
-- 5 2 10 NULL 10 ok
更新： Ed Harpers comment 真的让我重新思考我的方法。
所以第三次试炼来了。在这里，我首先搜索哪些行代表进入课程，哪些代表离开。然后我取所有结束时间的总和并减去所有开始时间的总和。我认为它更正确，但并不完美。
DECLARE @courseid INT;
SET @courseid = 1;
WITH numberedcte (rn, id, userid, courseid, sessionid, requestdate)
AS (
SELECT ROW_NUMBER() OVER (PARTITION BY sessionid, userid ORDER BY id)
, id
, userid
, courseid
, sessionid
, requestdate
FROM PageLogSample
)
, typedcte (rowtype, id, userid, courseid, sessionid, requestdate, nextrequestdate)
AS (
SELECT CASE
WHEN previousrequest.courseid = nextrequest.courseid
THEN 'between'
WHEN previousrequest.courseid IS NULL
OR nextrequest.courseid = numberedcte.courseid
THEN 'begin'
WHEN nextrequest.courseid IS NULL
OR previousrequest.courseid = numberedcte.courseid
THEN 'end'
ELSE 'error?'
END AS rowtype
, numberedcte.id
, numberedcte.userid
, numberedcte.courseid
, numberedcte.sessionid
, numberedcte.requestdate
, nextrequest.requestdate
FROM numberedcte
LEFT JOIN numberedcte previousrequest
ON previousrequest.userid = numberedcte.userid
AND previousrequest.sessionid = numberedcte.sessionid
AND previousrequest.rn = numberedcte.rn - 1
LEFT JOIN numberedcte nextrequest
ON nextrequest.userid = numberedcte.userid
AND nextrequest.sessionid = numberedcte.sessionid
AND nextrequest.rn = numberedcte.rn + 1
WHERE numberedcte.courseid = @courseid
AND (
nextrequest.courseid = @courseid
OR previousrequest.courseid = @courseid
)
)
, beginsum (userid, value)
AS (
SELECT userid, SUM(DATEPART(MINUTE, requestdate))
FROM typedcte
WHERE rowtype = 'begin'
GROUP BY userid
)
, endsum (userid, value)
AS (
SELECT userid, SUM(DATEPART(MINUTE, ISNULL(nextrequestdate, requestdate)))
FROM typedcte
WHERE rowtype = 'end'
GROUP BY userid
)
SELECT beginsum.userid
, endsum.value - beginsum.value AS duration
FROM beginsum
INNER JOIN endsum
ON beginsum.userid = endsum.userid
;
这里唯一的问题是我只能从原始样本数据中获得用户 1 和 5 的输出。添加的用户 6 也给出了正确的输出。添加的用户 7 现在给了我一个令人满意的输出。用户 8 几乎是完美的，我从第一行到第二行错过了一分钟。
-- userid duration
-- 1 10
-- 5 10
-- 6 10
-- 7 9
-- 8 9
我觉得我离完全正确还有几英寸的距离。唯一缺少的持续时间来自未在组中发生的页面请求。有人可以帮我找到一种方法来获得孤独的网页浏览量吗？
更新： 这是第四次试炼。在这里，我为每个请求分配一个值并总结它们。它并没有给我希望的输出，但看起来它可能已经足够好了。
DECLARE @courseid INT;
SET @courseid = 1;
WITH numberedcte (rn, userid, courseid, sessionid, requestdate)
AS (
SELECT ROW_NUMBER() OVER (PARTITION BY sessionid, userid ORDER BY id)
, userid
, courseid
, sessionid
, requestdate
FROM PageLogSample
)
, valuecte (value, userid, courseid, sessionid)
AS (
SELECT CASE
--alone
WHEN ( previousrequest.courseid IS NULL
OR previousrequest.courseid <> numberedcte.courseid
)
AND nextrequest.courseid <> numberedcte.courseid
THEN DATEDIFF(MINUTE, numberedcte.requestdate, nextrequest.requestdate)
--between
WHEN previousrequest.courseid = nextrequest.courseid
THEN 0
--begin
WHEN previousrequest.courseid IS NULL
OR nextrequest.courseid = numberedcte.courseid
THEN -1 * DATEPART(MINUTE, numberedcte.requestdate)
--ignored (end with no next request)
WHEN nextrequest.courseid IS NULL
AND previousrequest.courseid <> numberedcte.courseid
THEN 0
--end
WHEN nextrequest.courseid IS NULL
OR previousrequest.courseid = numberedcte.courseid
THEN DATEPART(MINUTE, ISNULL(nextrequest.requestdate, numberedcte.requestdate))
--impossible?
ELSE 0
END
, numberedcte.userid
, numberedcte.courseid
, numberedcte.sessionid
FROM numberedcte
LEFT JOIN numberedcte previousrequest
ON previousrequest.userid = numberedcte.userid
AND previousrequest.sessionid = numberedcte.sessionid
AND previousrequest.rn = numberedcte.rn - 1
LEFT JOIN numberedcte nextrequest
ON nextrequest.userid = numberedcte.userid
AND nextrequest.sessionid = numberedcte.sessionid
AND nextrequest.rn = numberedcte.rn + 1
WHERE numberedcte.courseid = @courseid
)
SELECT userid
, courseid
, COUNT(DISTINCT sessionid) AS sessioncount
, SUM(value) AS duration
FROM valuecte
GROUP BY userid
, courseid
ORDER BY userid
;
如您所见，结果并不完全符合我的预期。
-- userid courseid sessioncount duration
-- 1 1 1 10
-- 2 1 1 3
-- 3 1 1 6
-- 4 1 1 4
-- 5 1 2 10
-- 6 1 1 10
-- 7 1 1 9
-- 8 1 1 10
我的真实数据库的本地副本的性能很糟糕。因此，如果有人有想法以更高效的方式编写此内容......拍摄。
更新： 性能上去了。我添加了一个索引，它现在很有效。
【问题讨论】：
-
好问题 - 就脚本和示例而言，使其更容易解决。
-
数据的难点在于requestdate的含义不一致。有时是课程的开始时间，有时是课程的结束时间。
-
对这个好问题的好评论。
-
为了帮助我理解这个逻辑，如果我们改为查看 CourseID 2，您能否解释一下 UserID 3 的持续时间是多少？谢谢！
-
我感觉用户 3 在课程 2 上花了 9 分钟。(5 - 3) + (12 - 5) + (? - 15) = 9。问号表示我不知道如何很长一段时间他都在最后一页，因为没有后续的页面请求。也许我应该更好地解释它。这是服务器上的网页请求日志。每行都是一个页面请求。课程是网站的不同部分。我的任务是找出用户在每门课程上花费的时间。我需要替换以前的实现，因为它不够快。理想情况下，我的结果将与旧的实施结果相同。
标签： sql sql-server sql-server-2005 duration