计算某些记录不完整时登录和注销时间之间的持续时间

Question

我想计算用户在网站上花费的总时间。有3种情况。

存在用户登录时间和退出时间的记录。

-->总时间应该是登录和退出的时间差之和
有用户登录时间记录，但没有登出时间。

-->总时间应该标记为-1.
用户多次登录，只有1次退出

-->总时间应该是最早登录时间和退出时间的时间差之和。

我的table

CREATE TABLE #my_table
(
    id BIGINT                      IDENTITY PRIMARY KEY
    ,userID                        INT
    ,login_time                    DATETIME
    ,logout_time                   DATETIME
);

INSERT INTO #my_table
          SELECT 222222, '2016-05-19 01:06:00.000', '2016-05-19 01:10:00.000'
UNION ALL SELECT 222222, '2016-05-19 01:12:00.000', '2016-05-19 01:20:00.000'
UNION ALL SELECT 333333, '2016-05-24 14:44:00.000', '2016-05-24 14:47:00.000'
UNION ALL SELECT 333333, '2016-05-24 14:59:00.000', NULL
UNION ALL SELECT 444444, '2016-05-24 14:48:00.000', '2016-05-24 14:49:00.000'
UNION ALL SELECT 444444, '2016-05-24 14:50:00.000', NULL
UNION ALL SELECT 444444, '2016-05-24 14:51:00.000', NULL
UNION ALL SELECT 444444, '2016-05-24 14:53:00.000', '2016-05-24 14:59:00.000'

预期结果

对于大多数情况，数据库中捕获的记录将是情况1，但有时也会捕获情况2和情况3。我需要一个脚本来计算所有情况下的总登录时间。

如何查询？

Answer 1

IF OBJECT_ID('tempdb..#my_table') IS NOT NULL
    DROP TABLE #my_table

CREATE TABLE #my_table
(
    id BIGINT                      IDENTITY PRIMARY KEY
    ,userID                        INT
    ,login_time                    DATETIME
    ,logout_time                   DATETIME
);

DECLARE @MT TABLE
(
     id                            BIGINT
    ,userID                        INT
    ,login_time                    DATETIME
    ,logout_time                   DATETIME
);


DECLARE @DRes TABLE (
    userID          INT,
    logtime         INT
    )

DECLARE @Counter1 INT = 0,
    @login_time1        DATETIME,
    @logout_time1       DATETIME,
    @login_time2        DATETIME

INSERT INTO #my_table VALUES
     (222222, '2016-05-19 01:06:00.000', '2016-05-19 01:10:00.000')
    ,(222222, '2016-05-19 01:12:00.000', '2016-05-19 01:20:00.000')
    ,(333333, '2016-05-24 14:44:00.000', '2016-05-24 14:47:00.000')
    ,(333333, '2016-05-24 14:59:00.000', NULL)
    ,(444444, '2016-05-24 14:48:00.000', '2016-05-24 14:49:00.000')
    ,(444444, '2016-05-24 14:50:00.000', NULL)
    ,(444444, '2016-05-24 14:51:00.000', NULL)
    ,(444444, '2016-05-24 14:53:00.000', '2016-05-24 14:59:00.000')

INSERT INTO @MT
    SELECT * FROM #my_table

;WITH MaxLog 
    AS (
    SELECT userID, MAX(login_time) AS max_login
        FROM @MT
        GROUP BY userID
    ),

DelRec
    AS (
    SELECT ml.userID
        FROM MaxLog ml
        LEFT JOIN @MT mt
            ON ml.userID = mt.userID
        WHERE mt.logout_time IS NULL
            AND ml.max_login = mt.login_time
    )

DELETE mt
    FROM @MT mt
    INNER JOIN
        DelRec dr
        ON mt.userID = dr.userID
    WHERE mt.logout_time IS NOT NULL

;WITH StillIn
    AS (
    SELECT userID, COUNT(*) AS cnt
        FROM @MT
        GROUP BY userID
            HAVING COUNT(*) = 1
    )

UPDATE mt
    SET logout_time = DATEADD(mi,-1,login_time)
        FROM @MT mt
        JOIN StillIn si
            ON si.UserID = mt.UserId

WHILE @Counter1 < (SELECT MAX(id) FROM @MT)
    BEGIN
    SET @Counter1 += 1
    SET @login_time1 = (SELECT login_time FROM @MT WHERE id = @Counter1)
    SET @logout_time1 = (SELECT logout_time FROM @MT WHERE id = @Counter1)
    IF @logout_time1 IS NULL
        BEGIN
        IF @login_time2 IS NULL
            BEGIN
            SET @login_time2 = @login_time1
            END
        END
    ELSE
        BEGIN
        IF @login_time2 IS NULL
            BEGIN
            INSERT INTO @DRes
                SELECT userID, DATEDIFF(mi,@login_time1,@logout_time1)
                FROM @MT
                    WHERE id = @Counter1
            END
        ELSE
            BEGIN
            INSERT INTO @DRes
                SELECT userID, DATEDIFF(mi,@login_time2,@logout_time1)
                FROM @MT
                    WHERE id = @Counter1
            SET @login_time2 = NULL
            END
        END
    END

SELECT userID, SUM(logtime)
    FROM @DRes
    GROUP BY userID

Answer 2

下面的查询多次使用 ROW_NUMBER 函数来选择所需的行，并在 logout_time 为 NULL 时使用 LEAD 函数来选择 "look ahead"。 LEAD 自 SQL Server 2012 起可用。

运行逐步查询，逐个 CTE 并检查中间结果以了解其工作原理。

CTE_Groups 是经典的 gaps-and-islands 查询，用于标记 logout_time.

中具有连续 NULL 的行

CTE_RN 以这种方式将数字分配给行，logout_time 中连续的 NULL 得到连续的数字。此结果在 CTE_Fixed 中过滤，以仅获取每组 NULL 的第一行。如果 logout_time 为 NULL，则 LEAD 函数用于从下一行中选取一个值来生成 fixed_logout_time.

具有 NULL logout_time 的行和具有非 NULL logout_time 的下一行将一起列在 CTE_Fixed 中。我们只需要从这些对中选择一行。相同的方法 - 在 CTE_FixedRN 中使用 ROW_NUMBER 并在 CTE_Sum.

中选择第一行

然后我们可以在分钟内计算出 Duration 并将总和分组 userID。

如果没有非 NULL logout_time，DATEDIFF 将 return NULL，它将被一些大的负数替换。在最后的 SELECT 中，负数 Duration 将被替换为 -1 以指示最后一个间隔仍然开放。

WITH
CTE_Groups
AS
(
    SELECT
        userID
        ,login_time
        ,logout_time
        ,ROW_NUMBER() 
            OVER(PARTITION BY userID ORDER BY login_time)
        - ROW_NUMBER() 
            OVER(PARTITION BY userID, logout_time ORDER BY login_time) AS GroupNumber
    FROM #my_table
)
,CTE_RN
AS
(
    SELECT
        userID
        ,login_time
        ,logout_time
        ,ROW_NUMBER()
            OVER(PARTITION BY userID, GroupNumber ORDER BY login_time) AS rn
    FROM CTE_Groups
)
,CTE_Fixed
AS
(
    SELECT
        userID
        ,login_time
        ,ISNULL(logout_time, LEAD(logout_time) 
            OVER(PARTITION BY userID ORDER BY login_time)) AS fixed_logout_time
    FROM CTE_RN
    WHERE rn = 1
)
,CTE_FixedRN
AS
(
    SELECT
        userID
        ,login_time
        ,fixed_logout_time
        ,ROW_NUMBER() 
            OVER(PARTITION BY userID, fixed_logout_time ORDER BY login_time) AS rn
    FROM CTE_Fixed
)
,CTE_Sum
AS
(
    SELECT
        userID
        ,SUM(ISNULL(
            DATEDIFF(minute, login_time, fixed_logout_time),
            -1000000)) AS Duration
    FROM CTE_FixedRN
    WHERE rn = 1
    GROUP BY userID
)
SELECT
    userID
    ,CASE WHEN Duration < 0 THEN -1 ELSE Duration END AS Duration
FROM CTE_Sum
ORDER BY userID;

结果

+--------+----------+
| userID | Duration |
+--------+----------+
| 222222 |       12 |
| 333333 |       -1 |
| 444444 |       10 |
+--------+----------+

计算某些记录不完整时登录和注销时间之间的持续时间

Calculate duration between login and logout time when some records are incomplete

sql

sql-server

tsql

intervals

date-difference