查找 SQL Server 2008 中用户的日期范围之间的差距

Question

我需要知道用户登录了多少时间（天）以及未登录时间之间的间隔。在此 table 中，我仅存储 ID 及其开始和结束日期（分别为 ID、INI、FIN）。我已经设法通过将用户与行号分组然后将最新日志与以下日志进行比较等来基于条件检测三个记录上的差距。

问题是我曾经有过 n 条日志的人，我不能写 n 条左连接和 n 条条件。我希望使我当前的代码更具可扩展性，以更递归和更多用户的方式检测这些差距 "understandable"。

    CREATE TABLE [dbo].[baseRecurrentes](
    [ID] [nvarchar](8) NULL,
    [INI] [datetime] NULL,
    [FIN] [datetime] NULL
) ON [PRIMARY]
GO
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA1', CAST(0x0000A9C800000000 AS DateTime), CAST(0x0000A9E600000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA1', CAST(0x0000A9E700000000 AS DateTime), CAST(0x0000AA0200000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA1', CAST(0x0000AA0300000000 AS DateTime), CAST(0x0000AA2100000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA1', CAST(0x0000AA2200000000 AS DateTime), CAST(0x0000AA3F00000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA2', CAST(0x0000A9D600000000 AS DateTime), CAST(0x0000A9D900000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA2', CAST(0x0000A9EB00000000 AS DateTime), CAST(0x0000A9ED00000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA2', CAST(0x0000A9F000000000 AS DateTime), CAST(0x0000A9F100000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA3', CAST(0x0000AA1A00000000 AS DateTime), CAST(0x0000AA5A00000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA4', CAST(0x0000A9CA00000000 AS DateTime), CAST(0x0000A9CB00000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000A8DC00000000 AS DateTime), CAST(0x0000A8F100000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000A8F200000000 AS DateTime), CAST(0x0000A90F00000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000A91000000000 AS DateTime), CAST(0x0000A92E00000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000A92F00000000 AS DateTime), CAST(0x0000A94D00000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000A94E00000000 AS DateTime), CAST(0x0000A96B00000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000A96C00000000 AS DateTime), CAST(0x0000A98A00000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000A98B00000000 AS DateTime), CAST(0x0000A9A800000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000A9A900000000 AS DateTime), CAST(0x0000A9C700000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000A9C800000000 AS DateTime), CAST(0x0000A87900000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000A9E700000000 AS DateTime), CAST(0x0000AA0200000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000AA0300000000 AS DateTime), CAST(0x0000AA2100000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000AA2200000000 AS DateTime), CAST(0x0000AA3F00000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA5', CAST(0x0000AA4000000000 AS DateTime), CAST(0x0000AA5000000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA6', CAST(0x0000AA0900000000 AS DateTime), CAST(0x0000AA2900000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA7', CAST(0x0000A96C00000000 AS DateTime), CAST(0x0000A98A00000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA7', CAST(0x0000A98B00000000 AS DateTime), CAST(0x0000A9A800000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA7', CAST(0x0000A9A900000000 AS DateTime), CAST(0x0000A9C700000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA7', CAST(0x0000A85B00000000 AS DateTime), CAST(0x0000A87900000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA7', CAST(0x0000A9E700000000 AS DateTime), CAST(0x0000AA0200000000 AS DateTime))
INSERT [dbo].[baseRecurrentes] ([ID], [INI], [FIN]) VALUES (N'PERSONA7', CAST(0x0000AA0300000000 AS DateTime), CAST(0x0000AA2100000000 AS DateTime))

;WITH CTE AS (
    SELECT *, RN = ROW_NUMBER()OVER(PARTITION BY ID ORDER BY INI DESC)
    FROM BASERECURRENTES
), CTE2 AS (
    --
    SELECT DISTINCT T.ID,
    -- 
    CAST(T2.INI AS DATETIME) AS INI1, CAST(T2.FIN AS DATETIME) AS FIN1, 
    -- 
    CAST(T3.INI AS DATETIME) AS INI2, CAST(T3.FIN AS DATETIME) AS FIN2, 
    --
    CAST(T4.INI AS DATETIME) AS INI3, CAST(T4.FIN AS DATETIME) AS FIN3 
    --
    FROM CTE T
    LEFT JOIN CTE T2 ON T.ID = T2.ID AND T2.RN = 1 
    LEFT JOIN CTE T3 ON T.ID = T3.ID AND T3.RN = 2
    LEFT JOIN CTE T4 ON T.ID = T4.ID AND T4.RN = 3

), CTE3 AS (
    SELECT *, MSG = (CASE 
                                --NO GAPS ON 3 LOGS
                                WHEN (INI1 - 1 BETWEEN INI2 AND FIN2) AND (INI2 - 1 BETWEEN INI3 AND FIN3) THEN 'SEC2' 
                                --NO GAPS ON 2 LOGS
                                WHEN (INI1 - 1 BETWEEN INI2 AND FIN2) THEN 'SEC1' 
                                --NO GAP AT ALL
                                ELSE 'NO SEC'
                            END)

    FROM CTE2
)
SELECT * FROM CTE3
ORDER BY ID ASC

我希望 table 显示用户 ID，"gap days"（未记录时间的总和）和显示差距所在的消息。

ID       GD  MSG
-------------------
PERSONA2 5   GAP ON X-Y

Answer 1

我认为这会通过将每个用户的跨度与用户之前的跨度进行比较，为您提供您正在寻找的 "recursive" 结果。这向用户总结了所有结果；从 CTE2 中选择可以 return 有间隙或 Ini/Fin 的跨度被反转。

;WITH CTE AS (
    SELECT *,
        Person_RN = ROW_NUMBER() OVER (PARTITION BY ID ORDER BY INI, FIN),
        Person_Count = COUNT(*) OVER (PARTITION BY ID)
    FROM BASERECURRENTES
), CTE2 AS (
    SELECT T.*,
        N_Days_Logged_On = 
        CASE 
            WHEN DATEDIFF(DAY, T.Ini, T.Fin) < 0 THEN NULL -- Reversed span
            ELSE DATEDIFF(DAY, T.Ini, T.Fin) 
        END,
        N_Days_Gap_From_Prev = 
        CASE
            WHEN T.Person_RN = 1 THEN NULL -- First span
            WHEN DATEADD(DAY, 1, tPrev.Fin) = t.Ini THEN 0 -- OP allowed a one-day gap
            ELSE DATEDIFF(DAY, tPrev.Fin, T.Ini) 
        END,
        tPrev.Ini AS Ini_Prev, tPrev.Fin AS Fin_Prev
    FROM CTE T
    LEFT JOIN CTE TPrev ON T.ID = TPrev.ID AND TPrev.Person_RN = (T.Person_RN - 1)
    -- LEFT JOIN CTE TNext ON T.ID = TNext.ID AND TNext.Person_RN = (T.Person_RN + 1) 
), CTE3 AS (
    SELECT ID,
        GD = SUM(N_Days_Gap_From_Prev),
        N_Days_Logged_On = SUM(N_Days_Logged_On),
        N_Logs = MAX(Person_Count),
        N_Spans_WO_Gaps = COUNT(CASE WHEN N_Days_Gap_From_Prev = 0 THEN ID ELSE NULL END),
        N_Spans_W_Gaps = COUNT(NULLIF(N_Days_Gap_From_Prev, 0)),
        -- Captures reversed/invalid(?) spans, plus incomplete spans (null Ini or Fin)
        N_Spans_Suspect = COUNT(CASE WHEN N_Days_Logged_On IS NULL THEN ID ELSE NULL END)
    FROM CTE2
    GROUP BY ID
)
SELECT *
FROM CTE3
ORDER BY ID

我对此做了一些假设。从您的示例来看，logon/logoff 似乎是存储为日期时间的日期，但时间部分不是问题（例如，在 00:00 注销与 23:59 相同）。正如您在 CTE2 中看到的那样，DATEDIFF 结果不一定直观（例如，2019-01-01 到 2019-01-31 的 DATEDIFF 是 30 天）。

根据示例数据，这个 returns：

PERSONA1：4 条日志，0 天间隔，116 天登录，0 间隔间隔
PERSONA2：3个日志，21个gap days，6天登录，所有span都有gap。

Answer 2

您可以使用您的 ROW_NUMBER() 加入相同的 CTE 并寻找差距。这是我的建议：

;WITH CTE AS (
    SELECT *, RN = ROW_NUMBER()OVER(PARTITION BY ID ORDER BY INI DESC)
    FROM BASERECURRENTES
), CTE2 AS (
    SELECT * 
         , (SELECT TOP 1 FIN FROM CTE b WHERE a.ID = b.ID and b.RN > a.RN) ENDOFLAST
    FROM CTE a
), CTE3 AS (
    SELECT *
         , DATEDIFF(DAY, INI, ENDOFLAST) GAPDAYS
    FROM CTE2
)
SELECT *
FROM CTE3
WHERE GAPDAYS < -1
ORDER BY ID ASC

和 link 到 SQL Fiddle。我没有完全理解实际消息，但我认为结果集包含创建消息所需的一切。

查找 SQL Server 2008 中用户的日期范围之间的差距

Find gaps between date ranges for users in SQL Server 2008

sql

sql-server

common-table-expression

sql-server-2008