获取 SQL Server 上员工的重新分配时间(以天为单位)

Question

我有一个分配表，我需要计算员工完成旧工作后的重新分配时间(以天为单位(。

示例数据如下：

DROP TABLE #temptable
CREATE TABLE #temptable 
(
[EmployeeID] INT, 
[CustomerID] VARCHAR(255),
[Startdate] SMALLDATETIME,
[Enddate] SMALLDATETIME, 
[AssignmentID] VARCHAR(255) 
)
INSERT INTO #temptable
VALUES
( 260, '75569', N'2019-06-17T00:00:00', N'2019-06-17T00:00:00', '1139597' ), 
( 365, '77234', N'2019-08-12T00:00:00', NULL, '1145495' ), 
( 365, '77234', N'2019-08-19T00:00:00', NULL, '1146052' ), 
( 531, '82224', N'2019-07-12T00:00:00', N'2019-07-16T00:00:00', '1142003' ), 
( 531, '73793', N'2019-07-07T00:00:00', N'2019-07-15T00:00:00', '1141124' ), 
( 567, '41965', N'2019-07-08T00:00:00', N'2019-07-08T00:00:00', '1141186' ), 
( 567, '73793', N'2019-07-05T00:00:00', N'2019-07-07T00:00:00', '1140546' ), 
( 626, '73720', N'2019-08-07T00:00:00', N'2019-08-12T00:00:00', '1144431' ), 
( 718, '94251', N'2019-07-18T00:00:00', N'2019-07-17T00:00:00', '1142617' ), 
( 718, '20116', N'2019-08-12T00:00:00', N'2019-08-12T00:00:00', '1144671' ), 
( 718, '20116', N'2019-07-18T00:00:00', N'2019-07-19T00:00:00', '1142808' ), 
( 759, '93619', N'2019-07-30T00:00:00', N'2019-07-30T00:00:00', '1143923' ), 
( 780, '89916', N'2019-07-15T00:00:00', N'2019-07-15T00:00:00', '1142101' ), 
( 780, '89916', N'2019-07-15T00:00:00', N'2019-07-25T00:00:00', '1142249' ), 
( 781, '65921', N'2019-09-16T00:00:00', N'2019-09-24T00:00:00', '1147642' ), 
( 933, '67299', N'2019-07-02T00:00:00', N'2019-07-02T00:00:00', '1140805' ), 
( 936, '93619', N'2019-07-21T00:00:00', N'2019-07-22T00:00:00', '1143013' ), 
( 1056, '71469', N'2019-09-25T00:00:00', NULL, '1148611' ), 
( 1114, '14617', N'2019-07-16T00:00:00', N'2019-07-25T00:00:00', '1143005' ), 
( 1262, '73793', N'2019-09-16T00:00:00', N'2019-09-16T00:00:00', '1147541' ), 
( 1262, '93626', N'2019-07-06T00:00:00', N'2019-07-07T00:00:00', '1141098' ), 
( 1262, '93626', N'2019-09-14T00:00:00', N'2019-09-14T00:00:00', '1147538' ), 
( 1920, '75569', N'2019-06-17T00:00:00', N'2019-06-26T00:00:00', '1139589' ), 
( 2019, '93626', N'2019-09-07T00:00:00', N'2019-09-07T00:00:00', '1147220' ), 
( 2248, '93626', N'2019-08-31T00:00:00', N'2019-09-01T00:00:00', '1146398' ), 
( 2248, '93626', N'2019-08-24T00:00:00', N'2019-08-25T00:00:00', '1145809' ), 
( 2248, '93626', N'2019-09-07T00:00:00', N'2019-09-08T00:00:00', '1146865' ), 
( 2248, '93626', N'2019-09-14T00:00:00', N'2019-09-15T00:00:00', '1147507' ), 
( 2772, '177', N'2019-07-08T00:00:00', N'2019-07-29T00:00:00', '1141670' ), 
( 2772, '146', N'2019-06-20T00:00:00', N'2019-06-30T00:00:00', '1139511' ), 
( 2772, '90730', N'2019-07-20T00:00:00', N'2019-07-27T00:00:00', '1142917' ), 
( 2772, '55154', N'2019-08-16T00:00:00', N'2019-08-16T00:00:00', '1145276' ), 
( 2772, '55154', N'2019-07-27T00:00:00', N'2019-07-27T00:00:00', '1143509' ), 
( 2772, '77234', N'2019-07-30T00:00:00', N'2019-08-11T00:00:00', '1144407' ), 
( 3026, '73793', N'2019-09-24T00:00:00', N'2019-09-24T00:00:00', '1148379' ), 
( 3026, '93626', N'2019-09-13T00:00:00', N'2019-09-15T00:00:00', '1147492' ), 
( 3158, '82228', N'2019-08-19T00:00:00', N'2019-08-22T00:00:00', '1145372' ), 
( 3158, '84283', N'2019-08-13T00:00:00', N'2019-08-16T00:00:00', '1144849' ), 
( 3422, '41965', N'2019-08-21T00:00:00', N'2019-09-17T00:00:00', '1145629' ), 
( 3666, '93626', N'2019-07-26T00:00:00', N'2019-07-28T00:00:00', '1143451' ), 
( 3666, '93626', N'2019-07-19T00:00:00', N'2019-07-21T00:00:00', '1142714' ),
( 100, '86985', N'2019-09-23T00:00:00', N'2019-09-26T00:00:00' , '11427d14' ), 
( 100, '86985', N'2019-09-23T00:00:00', N'2019-09-26T00:00:00'  ,'11427d14' )

我使用了下面的脚本来获取开始日期和结束日期之间的天数差异。

要求：我需要为不同客户获得员工的重新分配天数。

SELECT 
EmployeeId,
CustomerId,
StartDate, EndDate,
DATEDIFF(DAY, A, StartDate) AS duration
FROM
(SELECT 
*,
LAG(EndDate) OVER (PARTITION BY EmployeeId, CustomerId ORDER BY StartDate) AS A
FROM #temptable
WHERE EndDate IS NOT NULL) b
ORDER BY 1 

在我的示例数据中：

• 对于EmployeeId = 3666，不同的客户有两个任务，但我没有得到两个任务之间的持续时间。

• 对于EmployeeId = 100，它是两个分配的同一客户，所以我不想显示 -3 值。

Answer 1

试试： 如果联接计数不够，可以增加联接

select distinct 
t.EmployeeID,   t.CustomerID    ,   t.AssignmentID,
isnull(DATEDIFF(DAY,t.Startdate,t.Enddate),0)+
isnull(DATEDIFF(DAY,t1.Startdate,t1.Enddate),0)+
isnull(DATEDIFF(DAY,t2.Startdate,t2.Enddate),0)+
isnull(DATEDIFF(DAY,t3.Startdate,t3.Enddate),0)+
isnull(DATEDIFF(DAY,t4.Startdate,t4.Enddate),0) sum FROM #temptable t
left join #temptable t1 on t.EmployeeID = t1.EmployeeID and t.AssignmentID  != t1.AssignmentID
left join #temptable t2 on t.EmployeeID = t2.EmployeeID and t2.AssignmentID != t1.AssignmentID and t2.AssignmentID != t.AssignmentID
left join #temptable t3 on t.EmployeeID = t3.EmployeeID and t3.AssignmentID != t2.AssignmentID and t3.AssignmentID != t1.AssignmentID and t3.AssignmentID != t.AssignmentID
left join #temptable t4 on t.EmployeeID = t4.EmployeeID and t4.AssignmentID != t3.AssignmentID and t4.AssignmentID != t2.AssignmentID and t4.AssignmentID != t.AssignmentID
order by t.EmployeeID