Sql 基于连续日期查找最近日期
我有一张表，列出了所有员工的缺勤（假期），我们想知道的是今天谁不在，以及他们回来的日期 不幸的是，缺勤没有ID，因此如果其中一个日期是今天，您不能仅从缺勤ID检索最大日期 但是，缺勤在输入时每天都会有一个递增的ID，因此我需要一个查询，如果有一个条目包含今天的日期，则会找到employeeID，然后递增缺勤ID列以查找缺勤的最大日期 表格示例（假设今天的日期为2014年11月11日，英国格式）： 因此，根据上述情况，我们希望返回日期为：Sql 基于连续日期查找最近日期,sql,sql-server,greatest-n-per-group,Sql,Sql Server,Greatest N Per Group,我有一张表，列出了所有员工的缺勤（假期），我们想知道的是今天谁不在，以及他们回来的日期 不幸的是，缺勤没有ID，因此如果其中一个日期是今天，您不能仅从缺勤ID检索最大日期 但是，缺勤在输入时每天都会有一个递增的ID，因此我需要一个查询，如果有一个条目包含今天的日期，则会找到employeeID，然后递增缺勤ID列以查找缺勤的最大日期 表格示例（假设今天的日期为2014年11月11日，英国格式）： 因此，根据上述情况，我们希望返回日期为： EmployeeID ReturnDate 10
EmployeeID ReturnDate
10 15/11/2014
21 12/11/2014
05 11/11/2014
大概我需要在每个条目上运行一个迭代子函数，其中包含employeeID匹配的今天的日期。因此，根据我相信您提出的问题，您希望根据您在系统中记录的假期返回一份今天休假的人员列表，以及他们预计何时返回的列表，只能在连续几天内工作 架构设置：
CREATE TABLE EmployeeAbsence
([AbsenceID] int, [EmployeeID] int, [AbsenceDate] DATETIME)
;
INSERT INTO EmployeeAbsence
([AbsenceID], [EmployeeID], [AbsenceDate])
VALUES
(100, 10, '2014-11-11'),
(101, 10, '2014-11-12'),
(102, 10, '2014-11-13'),
(103, 10, '2014-11-14'),
(104, 10, '2014-11-15'),
(107, 21, '2014-11-11'),
(108, 21, '2014-11-12'),
(120, 05, '2014-11-11'),
(130, 15, '2014-11-20')
;
;WITH cte AS (
SELECT EmployeeID, AbsenceDate
FROM dbo.EmployeeAbsence
WHERE AbsenceDate = CAST(GETDATE() AS DATE)
UNION ALL
SELECT e.EmployeeID, e.AbsenceDate
FROM cte
INNER JOIN dbo.EmployeeAbsence e ON e.EmployeeID = cte.EmployeeID
AND e.AbsenceDate = DATEADD(d,1,cte.AbsenceDate)
)
SELECT cte.EmployeeID, MAX(cte.AbsenceDate)
FROM cte
GROUP BY cte.EmployeeID
| EMPLOYEEID | Return Date |
|------------|---------------------------------|
| 5 | November, 11 2014 00:00:00+0000 |
| 10 | November, 15 2014 00:00:00+0000 |
| 21 | November, 12 2014 00:00:00+0000 |
CREATE TABLE EmployeeAbsence
([AbsenceID] int, [EmployeeID] int, [AbsenceDate] DATETIME)
;
INSERT INTO EmployeeAbsence
([AbsenceID], [EmployeeID], [AbsenceDate])
VALUES
(100, 10, '2014-11-11'),
(101, 10, '2014-11-12'),
(102, 10, '2014-11-13'),
(103, 10, '2014-11-14'),
(104, 10, '2014-11-15'),
(107, 21, '2014-11-11'),
(108, 21, '2014-11-12'),
(120, 05, '2014-11-11'),
(130, 15, '2014-11-20')
;
;WITH cte AS (
SELECT EmployeeID, AbsenceDate
FROM dbo.EmployeeAbsence
WHERE AbsenceDate = CAST(GETDATE() AS DATE)
UNION ALL
SELECT e.EmployeeID, e.AbsenceDate
FROM cte
INNER JOIN dbo.EmployeeAbsence e ON e.EmployeeID = cte.EmployeeID
AND e.AbsenceDate = DATEADD(d,1,cte.AbsenceDate)
)
SELECT cte.EmployeeID, MAX(cte.AbsenceDate)
FROM cte
GROUP BY cte.EmployeeID
| EMPLOYEEID | Return Date |
|------------|---------------------------------|
| 5 | November, 11 2014 00:00:00+0000 |
| 10 | November, 15 2014 00:00:00+0000 |
| 21 | November, 12 2014 00:00:00+0000 |
CREATE TABLE EmployeeAbsence
([AbsenceID] int, [EmployeeID] int, [AbsenceDate] DATETIME)
;
INSERT INTO EmployeeAbsence
([AbsenceID], [EmployeeID], [AbsenceDate])
VALUES
(100, 10, '2014-11-11'),
(101, 10, '2014-11-12'),
(102, 10, '2014-11-13'),
(103, 10, '2014-11-14'),
(104, 10, '2014-11-15'),
(107, 21, '2014-11-11'),
(108, 21, '2014-11-12'),
(120, 05, '2014-11-11'),
(130, 15, '2014-11-20')
;
;WITH cte AS (
SELECT EmployeeID, AbsenceDate
FROM dbo.EmployeeAbsence
WHERE AbsenceDate = CAST(GETDATE() AS DATE)
UNION ALL
SELECT e.EmployeeID, e.AbsenceDate
FROM cte
INNER JOIN dbo.EmployeeAbsence e ON e.EmployeeID = cte.EmployeeID
AND e.AbsenceDate = DATEADD(d,1,cte.AbsenceDate)
)
SELECT cte.EmployeeID, MAX(cte.AbsenceDate)
FROM cte
GROUP BY cte.EmployeeID
| EMPLOYEEID | Return Date |
|------------|---------------------------------|
| 5 | November, 11 2014 00:00:00+0000 |
| 10 | November, 15 2014 00:00:00+0000 |
| 21 | November, 12 2014 00:00:00+0000 |
选择WHERE AbsenceDate = CAST(GETDATE() AS DATE)
EmployeeID缺勤日期employeedisease-- add a day to the cte.AbsenceDate from the first SELECT
e.AbsenceDate = DATEADD(d,1,cte.AbsenceDate)
SELECTMAXSELECT cte.EmployeeID, MAX(cte.AbsenceDate)
FROM cte
GROUP BY cte.EmployeeID
内部连接的修改应该排除周末，如果它检测到增加一天将导致星期六，则增加额外的天数：
INNER JOIN dbo.EmployeeAbsence e ON e.EmployeeID = cte.EmployeeID
AND e.AbsenceDate = CASE WHEN datepart(dw,DATEADD(d,1,cte.AbsenceDate)) = 7
THEN DATEADD(d,3,cte.AbsenceDate)
ELSE DATEADD(d,1,cte.AbsenceDate) END
因此，当您添加一天时：datepart（dw，DATEADD（d，1，cte.AbsenceDate））=7
，如果结果是星期六（7），那么您需要添加3天而不是1天来获得星期一：DATEADD（d，3，cte.AbsenceDate）
您需要做一些事情来将此数据转换为可用格式。您需要能够确定团队的开始和结束位置。这个示例很难做到这一点，因为没有直接的分组列
为了计算组的开始和结束时间，您需要创建一个包含所有列的CTE，并使用LAG（）
从每一行的前一行中获取AbsenceID
和EmployeeID
。在这个CTE中，您还应该同时使用行编号（）
，这样我们就可以将行重新排序为相同的顺序
比如：
WITH
[AbsenceStage] AS (
SELECT [AbsenceID], [EmployeeID], [AbsenceDate]
,[RN] = ROW_NUMBER() OVER (ORDER BY [EmployeeID] ASC, [AbsenceDate] ASC, [AbsenceID] ASC)
,[AbsenceID_Prev] = LAG([AbsenceID]) OVER (ORDER BY [EmployeeID] ASC, [AbsenceDate] ASC, [AbsenceID] ASC)
,[EmployeeID_Prev] = LAG([EmployeeID]) OVER (ORDER BY [EmployeeID] ASC, [AbsenceDate] ASC, [AbsenceID] ASC)
FROM [HR_Absence]
)
[EmployeeID_Prev] IS NULL -- We have a new group if the previous row is null
OR [EmployeeID_Prev] <> [EmployeeID] -- Or if the previous row is for a different employee
OR [AbsenceID_Prev] <> ([AbsenceID]-1) -- Or if the AbsenceID is not sequential
....
FROM [AbsenceStage] AS [Row]
INNER JOIN [AbsenceStage] AS [First]
ON ([First].[RN] = (
-- Get the first row before ([RN] Less that or equal to) this one where it is the start of a grouping
SELECT MAX([RN]) FROM [AbsenceStage]
WHERE [RN] <= [Row].[RN] AND (
[EmployeeID_Prev] IS NULL
OR [EmployeeID_Prev] <> [EmployeeID]
OR [AbsenceID_Prev] <> ([AbsenceID]-1)
)
))
...
现在我们有了这些，我们可以将每一行与前一行进行比较，以查看当前行是否与前一行处于不同的“组”中
这种情况类似于：
WITH
[AbsenceStage] AS (
SELECT [AbsenceID], [EmployeeID], [AbsenceDate]
,[RN] = ROW_NUMBER() OVER (ORDER BY [EmployeeID] ASC, [AbsenceDate] ASC, [AbsenceID] ASC)
,[AbsenceID_Prev] = LAG([AbsenceID]) OVER (ORDER BY [EmployeeID] ASC, [AbsenceDate] ASC, [AbsenceID] ASC)
,[EmployeeID_Prev] = LAG([EmployeeID]) OVER (ORDER BY [EmployeeID] ASC, [AbsenceDate] ASC, [AbsenceID] ASC)
FROM [HR_Absence]
)
[EmployeeID_Prev] IS NULL -- We have a new group if the previous row is null
OR [EmployeeID_Prev] <> [EmployeeID] -- Or if the previous row is for a different employee
OR [AbsenceID_Prev] <> ([AbsenceID]-1) -- Or if the AbsenceID is not sequential
....
FROM [AbsenceStage] AS [Row]
INNER JOIN [AbsenceStage] AS [First]
ON ([First].[RN] = (
-- Get the first row before ([RN] Less that or equal to) this one where it is the start of a grouping
SELECT MAX([RN]) FROM [AbsenceStage]
WHERE [RN] <= [Row].[RN] AND (
[EmployeeID_Prev] IS NULL
OR [EmployeeID_Prev] <> [EmployeeID]
OR [AbsenceID_Prev] <> ([AbsenceID]-1)
)
))
...
然后，您可以将所有这些内容放到一个视图中，该视图为您提供员工ID
，以及每次缺勤的开始和结束日期。然后，您可以使用以下工具轻松地提取员工当前的休假：
WHERE CAST(CURRENT_TIMESTAMP AS date) BETWEEN [Absence_Begin] AND [Absence_End]
就像这里的另一个答案一样，我将创建休假间隔，但使用不同的方法。首先是守则：
declare @today date = getdate(); --use whatever date here
with g as (
select *, dateadd(day, -1 * row_number() over (partition by employeeid order by absencedate), AbsenceDate) as group_number
from employeeabsence
) , leave_intervals as (
select employeeid, min(absencedate) as [start], max(absencedate) as [end]
from g
group by EmployeeID, group_number
)
select employeeid, [start], [end]
from leave_intervals
where @today between [start] and [end]
作为解释，我们首先将日期值放入变量中。我选择了今天，但此代码将适用于传入的任何日期。接下来，我们创建一个公共表表达式（CTE），它将在表中添加一个分组列。这是溶液的肉，所以需要一些处理。在给定的时间间隔内，缺勤日期以每行一天的速度增加<代码>行数（）行数（）