计算多行SQL之间的日期差

Question

我需要计算多行之间的日期差异。场景是我有一辆可以在整个月以及当车辆被分配到不同项目时进行检查的车辆。我想计算每月或上个月将车辆分配给项目的天数。我已经尝试了多种方法，但我无法靠近。我对堆栈溢出比较陌生。如果有任何遗漏，我们深表歉意。请让我知道这是否可以做到。谢谢。

如果有帮助，所有列都在一个表中。请让我知道有关如何实现此目的的查询

我使用的是 SQL Server 2017。

原始数据

预期输出

Answer 1

我对这个解决方案并不感到自豪，但我认为它对你有用。我的方法是创建一个天数表，然后查看车辆每天分配给哪个项目。最后，按月和年汇总得到结果。我必须以脚本的形式执行此操作，因为您可以在递归 CTE 的定义中使用聚合函数，但您可能会找到一种不需要递归 CTE 的方法。

我创建了一个表变量来导入您的数据，所以我可以写这个。请注意，我为跨越数月的测试作业添加了一个额外的作业。

DECLARE @Vehicles AS TABLE
(
    [VehicleID]      INT     NOT NULL,
    [Project]        CHAR(2) NOT NULL,
    [InspectionDate] DATE    NOT NULL
);

INSERT INTO @Vehicles
(
    [VehicleID],
    [Project],
    [InspectionDate]
)
VALUES
(1, 'P1', '2021-08-20'),
(1, 'P1', '2021-09-05'),
(1, 'P2', '2021-09-15'),
(1, 'P3', '2021-09-20'),
(1, 'P2', '2021-10-10'),
(1, 'P1', '2021-10-20'),
(1, 'P3', '2021-10-21'),
(1, 'P2', '2021-10-22'),
(1, 'P4', '2021-11-15'),
(1, 'P4', '2021-11-25'),
(1, 'P4', '2021-11-30'),
(1, 'P1', '2022-02-05');

DECLARE @StartDate AS DATE, @EndDate AS DATE;

SELECT @StartDate = MIN([InspectionDate]), @EndDate = MAX([InspectionDate])
FROM   @Vehicles;

;WITH [seq]([n])
AS (SELECT 0 AS [n]
    UNION ALL
    SELECT [n] + 1
    FROM   [seq]
    WHERE  [n] < DATEDIFF(DAY, @StartDate, @EndDate)),
      [days]
AS (SELECT DATEADD(DAY, [n], @StartDate) AS [d]
    FROM   [seq]),
      [inspections]
AS (SELECT [VehicleID],
           [Project],
           [InspectionDate],
           LEAD([InspectionDate], 1) OVER (PARTITION BY [VehicleID]
                                           ORDER BY [InspectionDate]
                                          ) AS [NextInspectionDate]
    FROM   @Vehicles),
      [assignmentsByDay]
AS (SELECT [d].[d], [i].[VehicleID], [i].[Project]
    FROM   [days] AS [d]
           INNER JOIN [inspections] AS [i]
               ON [d].[d] >= [i].[InspectionDate]
                  AND [d] < [i].[NextInspectionDate])

SELECT   [assignmentsByDay].[VehicleID],
         [assignmentsByDay].[Project],
         MONTH([assignmentsByDay].[d]) AS [month],
         YEAR([assignmentsByDay].[d]) AS [year],
         COUNT(*) AS [daysAssigned]
FROM     [assignmentsByDay]
GROUP BY [assignmentsByDay].[VehicleID],
         [assignmentsByDay].[Project],
         MONTH([assignmentsByDay].[d]),
         YEAR([assignmentsByDay].[d])
ORDER BY [year], [month], [assignmentsByDay].[VehicleID], [assignmentsByDay].[Project]
OPTION(MAXRECURSION 0);

输出是：

VehicleID	Project	month	year	daysAssigned
1	P1	8	2021	12
1	P1	9	2021	14
1	P2	9	2021	5
1	P3	9	2021	11
1	P1	10	2021	1
1	P2	10	2021	20
1	P3	10	2021	10
1	P2	11	2021	14
1	P4	11	2021	16
1	P4	12	2021	31
1	P4	1	2022	31
1	P4	2	2022	4

Answer 2

我想你正在寻找这个：

select vehicleId
      , Project 
      , month(inspectiondate) month
      , year(inspectiondate) year
      , datediff(day , min(inspectiondate), case when max(inspectiondate) = min(inspectiondate) then eomonth(min(inspectiondate)) else max(inspectiondate) end) days      
from Vehicles
group by vehicleId, Project , month(inspectiondate), year(inspectiondate)

该查询在每个月/年中针对该项目中的每个特定车辆在该月/年中获取最大和最小检查日期并计算差异。

db小提琴here