我有一个表，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courseid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。
此处提供的数据应该导致课程1的每个用户持续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。这只能部分起作用。
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。
所以这是第三次试验。在这里，我首先搜索哪些行代表课程的入口，哪些代表某人离开。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也提供正确的输出。添加的用户7现在给我一个满意的输出。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
我觉得我距离完全正确的距离还有几英寸远。缺少的唯一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取孤独的综合浏览量吗？
更新 这是第四次试验。在这里，我为每个请求分配一个值并总结它们。它并没有给我提供我希望的输出，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
;
正如您所看到的，结果并非完全符合我的预期。
-- 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
在我的真实数据库的本地副本上，性能非常糟糕。因此，如果有人有想法以更高效的方式写这个......拍摄。
更新 表现上升。我添加了一个索引，它现在起了作用。
答案 0 :(得分：0)
抱歉，我认为您遇到了数据问题。查看提供的样本数据，用户2在1中持续12分钟，在2中持续2分钟。
您确定提供了正确的数据吗？
答案 1 :(得分：0)
这是我能得到的尽可能接近。用户ID 4失败。
正如我在评论中所说，requestdate有时是一个开始，有时是课程的结束，我看不出一个简单的一般规则来推导出它在给定行上扮演的角色。
DECLARE @courseid INT;
SET @courseid = 1;
WITH orderCTE
AS
(
SELECT *
,ROW_NUMBER() OVER (PARTITION BY sessionid
ORDER BY id
) AS rn
FROM PageLogSample
--order by rn
)
,startendCTE
AS
(
SELECT CASE WHEN start1.rn = 1
THEN start1.courseid
ELSE end1.courseid
END courseid
,start1.sessionid
,start1.userid
,DATEDIFF(mi,start1.requestdate,end1.requestdate) duration
FROM orderCTE AS start1
JOIN orderCTE AS end1
ON end1.rn = start1.rn + 1
AND end1.sessionid = start1.sessionid
)
SELECT courseid
,COUNT(1) sessionCount
,userid
,SUM(duration) totalDuration
FROM startendCTE
WHERE courseid = @courseid
GROUP BY courseid
,userid;
答案 2 :(得分：0)
这非常混乱，但它似乎适用于CourseID 1.我没有尝试过其他课程，所以你可能想测试一下！ ：d
基本前提是我正在获取目标CourseID的第一个和最后一个会话之间的持续时间，然后我减去任何不是指定CourseID但会话请求的会话的持续时间时间落在目标CourseID的最小和最大请求时间内。我希望这是有道理的。
绝对可以清除查询，可能使用CTE或其他东西。有趣的问题BTW！ ：）
DECLARE @courseid INT;
SET @courseid = 1;
SELECT
TargetCourse.UserID,
COUNT(Distinct(TargetCourse.SessionID)) as SessionCount,
SUM(TargetCourse.Duration - Coalesce(OtherCourses.Duration,0)) as Duration
FROM
(
SELECT
TargetCourse.UserID, TargetCourse.SessionID,
MIN(TargetCourse.RequestDate) FirstRequest, MAX(TargetCourse.RequestDate) LastRequest,
DATEDIFF(MINUTE, MIN(TargetCourse.RequestDate), MAX(TargetCourse.RequestDate)) AS duration
FROM
PageLogSample TargetCourse
WHERE
TargetCourse.CourseID = @courseid
GROUP BY
TargetCourse.UserID, TargetCourse.SessionID
) as TargetCourse
LEFT OUTER JOIN
(
SELECT
OtherCourses.UserID, OtherCourses.SessionID,
MIN(OtherCourses.RequestDate) AS FirstRequest, MAX(OtherCourses.RequestDate) AS LastRequest,
DATEDIFF(MINUTE, MIN(OtherCourses.RequestDate), MAX(OtherCourses.RequestDate)) AS duration
FROM
PageLogSample OtherCourses
WHERE
OtherCourses.CourseID <> @courseid AND
OtherCourses.RequestDate between
(Select MIN(RequestDate) From PageLogSample T Where T.UserID = OtherCourses.UserID and T.CourseID = @courseid) AND
(Select MAX(RequestDate) From PageLogSample T Where T.UserID = OtherCourses.UserID and T.CourseID = @courseid)
GROUP BY
OtherCourses.UserID, OtherCourses.SessionID
) as OtherCourses ON
OtherCourses.UserID = TargetCourse.UserID AND
OtherCourses.FirstRequest BETWEEN TargetCourse.FirstRequest and TargetCourse.LastRequest
Group By TargetCourse.UserID
答案 3 :(得分：0)
更多样本数据以及每个用户在每门课程中花费的时间的逻辑假设。
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, 3] = 3 minutes
-- there is no way to know how long the user was on that last page
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, 3] + [12, 15] = 6 minutes
-- the [5, 12] part was spent on a page of course 2
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, 3] + [13, 15] = 4 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, 5] + [7, 11] = 9 minutes
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'
-- there is nothing we can say about either of there requests
-- 0 minutes
UNION ALL SELECT 9, 1, 10, '00:10:00'
UNION ALL SELECT 9, 1, 11, '00:20:00'
;
现在我们得到这样的数据：
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 in session
WHEN previousrequest.courseid IS NULL
AND nextrequest.courseid IS NULL
THEN 0
--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
;
这是我得到的结果。我很满意。注意会话计数如何对用户9保持正确。
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
9 1 2 0
答案 4 :(得分：-1)
“数据是正确的，但很难从中得到相关的意义。”
我很想回应这是一个矛盾的术语。您不知道其含义的数据不是数据。
关于你原来的问题：
您需要的是一个DBMS，它为INTERVAL类型提供了不错的支持。没有SQL系统在那个联盟中发挥作用。除了一些教程系统之外，我自己的DBMS（在此上下文中没有进一步推动，所以没有链接）是我所知道的唯一提供此类问题所需的支持。
如果您有兴趣，请浏览“间隔类型”，“打包正常形式”，“时态数据”，最后您会遇到它。