如何计算重叠时间和冲突长度

Question

Employee 表有四列。员工，开始，结束，差异。 Diff 列是持续时间，计算方式为 End - Start。

我想找出开始和结束时间范围之间的冲突。 例如，员工 A 有三行： 第一行开始时间是 01:02，结束时间是 01:05，但第二行开始时间是 01:03，这是第一行数据的冲突。

样本数据：

   employee   StartDate    EndDate       Start         End        Diff 
    A       04/08/2019   04/08/2019       01:02:00   01:05:00       3
    A       04/08/2019   04/08/2019       01:03:00   01:08:00       5
    A       04/08/2019   04/08/2019       01:014:00  01:21:00       7
    B       04/08/2019   04/08/2019       02:00:00   02:17:00      17

我只想为员工 A 选择开始和结束时间重叠的特定开始和结束时间，并且想使用 t-sql 计算新列中的冲突总长度。我是新手，需要帮助。有人请吗？

SELECT TOP (100) a.ccx_employeename AS employee
    ,CONVERT(Date,[a].[ccx_starttime]) AS [Start Date],CONVERT(Date,[a].[ccx_endtime]) AS [End Date], CONVERT(time (0), a.ccx_starttime) AS StartTime
    , CONVERT(time (0), a.ccx_endtime) AS EndTime
    , CONVERT (time(0), (a.ccx_endtime - a.ccx_starttime)) AS Duration
FROM ccp_sim_MSCRM.dbo.Filteredccx_Recorded_Service as a 
where CONVERT(time (0), a.ccx_starttime) BETWEEN CONVERT(time (0), a.ccx_starttime) And CONVERT(time (0), a.ccx_endtime) 

由于第一行和第二行有冲突，我想显示这两行。在此示例中，冲突持续时间为 2 分钟。第一行结束时间是 01:05，但第二行开始时间是 01:03，所以冲突持续时间是 01:05 - 01:03 = 2 分钟

期望的输出

employee   StartDate    EndDate       Start         End        Diff 
A       04/08/2019   04/08/2019       01:02:00   01:05:00       3
A       04/08/2019   04/08/2019       01:03:00   01:08:00       5

duration of conflict : 2 mins

Answer 1

我会自己加入表格，虽然可能不是最有效的：

SELECT
    e1.employee,
    e1.Start as firstStart,
    e1.End as firstEnd,
    e2.Start as secondStart,
    e2.End as secondEnd,
    e1.End - e2.Start as conflictDuration
FROM
    Employee as e1 inner join
    Employee as e2 on (
        e1.employee = e2.employee and 
        e2.Start < e1.End and
        e2.End > e1.Start
    )

Answer 2

你的问题有几个部分：

• 查找冲突行
• 计算冲突时间
• 以您想要的格式输出

下面的解决方案仅涵盖前两部分，并假定一个组合的日期和时间字段。

我添加了一些唯一键来删除结果并“排序”行以进行比较。在下面的代码中，它是“id”。

declare @t table (id int identity,employee char(1), StartDateTime smalldatetime, EndDateTime smalldatetime, diff as DATEDIFF(minute,StartDateTime,EndDateTime))
insert into @t values('A','2019-04-08 01:02','2019-04-08 01:05')
insert into @t values('A','2019-04-08 01:03','2019-04-08 01:08')
insert into @t values('A','2019-04-08 01:14','2019-04-08 01:21')
insert into @t values('B','2019-04-08 02:00','2019-04-08 02:17')

SELECT T1.employee, T1.StartDateTime, T1.EndDateTime, T2.StartDateTime, T2.EndDateTime
,      (T1.diff + T2.diff)
       - DATEDIFF(minute, CASE WHEN T1.StartDateTime < T2.StartDateTime THEN T1.StartDateTime ELSE T2.StartDateTime END   -- MIN(Start)
                        , CASE WHEN T1.EndDateTime   > T2.EndDateTime   THEN T1.EndDateTime   ELSE T2.EndDateTime   END)  -- MAX(End)
       AS "duration of conflict"
FROM @t AS T1
JOIN @t AS T2
  ON  T2.employee = T1.employee
  AND T2.id       > T1.id        -- Each only once
  AND T2.StartDateTime < T1.EndDateTime
  AND T2.EndDateTime   > T1.StartDateTime

Answer 3

这对我来说是使用 LEAD/LAG 函数的理想场所。将其与一些子查询和IIF 语句结合起来，您就可以计算出您要查找的结果。

例子：

DECLARE @Employee TABLE
(
    Employee VARCHAR(1),
    startDate DATE,
    endDate DATE,
    [start] TIME,
    [end] TIME,
    diff AS DATEDIFF(MINUTE,[start],[end])
)

INSERT INTO @Employee (Employee, startDate, endDate, start, [end])
VALUES
    ('A',CAST('2019-04-08' AS DATE),CAST('2019-04-08' AS DATE),'01:02:00','01:05:00'),
    ('A',CAST('2019-04-08' AS DATE),CAST('2019-04-08' AS DATE),'01:03:00','01:08:00'),
    ('A',CAST('2019-04-08' AS DATE),CAST('2019-04-08' AS DATE),'01:14:00','01:21:00'),
    ('B',CAST('2019-04-08' AS DATE),CAST('2019-04-08' AS DATE),'02:00:00','02:17:00')

SELECT
    Employee.Employee,
    Employee.startDate,
    Employee.endDate,
    Employee.start,
    Employee.[end],
    diff,
    (IIF(ISNULL(lagConflict,0)>0,ISNULL(lagConflict,0),0)+IIF(ISNULL(Employee.leadConflict,0)>0,ISNULL(Employee.leadConflict,0),0)) AS conflict
FROM
    (
        SELECT
            Employee,
            startDate,
            endDate,
            start,
            [end],
            diff,
            DATEDIFF
            (
                MINUTE,
                [start],
                LAG([end],1)
                OVER
                (
                    PARTITION BY
                        Employee,
                        startDate,
                        endDate
                    ORDER BY
                        [start],
                        [end]
                )
            ) AS lagConflict,
            DATEDIFF
            (
                MINUTE,
                [end],
                LEAD([start],1)
                OVER
                (
                    PARTITION BY
                        Employee,
                        startDate,
                        endDate
                    ORDER BY
                        [start],
                        [end]
                )
            )*-1 AS leadConflict
        FROM
            @Employee
    ) AS Employee
WHERE
    Employee.leadConflict > 0
    OR Employee.lagConflict > 0;

Microsoft SQL 文档：LAG]1