sql - SQL查询从日志表中计算部分访问持续时间

Question

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每个用户在课程 1 中的持续时间为 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现错误。

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 的评论真的让我重新思考了我的方法。

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我的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： 这是第四次审判。在这里，我为每个请求分配一个值并总结它们。它并没有为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
;

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

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编写此内容......拍摄。

更新： 性能上升。我添加了一个索引，它现在很有魅力。

Answer 1

抱歉，我认为您的数据有问题。查看提供的样本数据，用户 2 在 courseid 1 中停留 12 分钟，在 courseid 2 中停留 2 分钟。

你确定你提供了正确的数据吗？

Answer 2

这是我能得到的最接近的。用户 ID 4 失败。

正如我在评论中所说，requestdate有时是课程的开始，有时是课程的结束，我看不到一个简单的一般规则来推导它在给定行中扮演的角色。

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;

Answer 3

这很混乱，但它似乎适用于 CourseID 1。我没有在其他课程中尝试过，所以你可能想测试一下！:D

基本前提是我得到目标 CourseID 的第一个会话和最后一个会话之间的持续时间，然后我减去不属于指定 CourseID 的任何会话的持续时间，但会话请求时间属于目标 CourseID 的最小和最大请求​​时间。我希望这是有道理的。

查询肯定可以被清理，可能使用 CTE 或其他东西。顺便说一句，有趣的问题！:)

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

Answer 4

更多示例数据和一个有希望的逻辑假设，即每个用户在每门课程中花费了多少时间。

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。请注意用户 9 的会话计数如何保持正确。

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

Answer 5

“数据是正确的，但很难从中得到相关的意义。”

我迫不及待地回应说这是一个自相矛盾的术语。您不知道其含义的数据不是数据。

至于你原来的问题：

您需要的是一个为 INTERVAL 类型提供良好支持的 DBMS。在那个联盟中没有 SQL 系统。除了一些教程系统之外，我自己的 DBMS（在这种情况下没有进一步推动，所以没有链接）是我所知道的唯一一个提供此类问题真正需要的支持的系统。

如果您有兴趣，可以在 Google 上搜索“区间类型”、“压缩范式”、“时间数据”，最终您会遇到。