SQL 查询从日志表计算部分访问持续时间

Question

我有一个表，用于记录每次加载网页时的用户 ID、类(class)、 session 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。

此处提供的数据应导致类(class) 1 中每个用户的持续时间为 10 分钟。我似乎无法理解这一点。

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'
;

首先尝试简单的方法。这会导致 session 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确实让我重新思考我的方法。

第三次审判来了。在这里，我首先搜索哪些行代表进入类(class)，哪些行代表离开。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。我认为这是比较正确的，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 也给出了正确的输出。添加的用户 7 现在给了我满意的输出。用户8几乎完美，从第一排到第二排我错过了一分钟。

-- userid  duration
-- 1       10
-- 5       10
-- 6       10
-- 7       9
-- 8       9

我觉得我距离完全正确还有几英寸的距离。唯一缺少的持续时间来自未成组发生的页面请求。有人可以帮我找到一种方法来获得孤独的综合浏览量吗？

更新:第四次审判来了。在这里，我为每个请求分配一个值并对它们进行总结。它没有给我确切的我希望的输出，但看起来它已经足够好了。

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......请拍摄。

更新:性能上去了。我添加了一个索引，现在它很有用了。

Answer 1

抱歉，我认为您遇到了数据问题。查看提供的示例数据，用户 2 在 courseid 1 中呆了 12 分钟，在 courseid 2 中呆了 2 分钟。

您确定提供了正确的数据吗？