周末病假统计

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英文:

Counting Sick days over the weekend

问题

我正在尝试解决以下(简化的)数据集中的问题:

姓名 日期 工作日 日历日 请假
人员A 2023-01-01 0 1 NULL
人员A 2023-01-07 0 1 NULL
人员A 2023-01-08 0 1 NULL
人员A 2023-01-13 1 1 病假
人员A 2023-01-14 0 1 NULL
人员A 2023-01-15 0 1 NULL
人员A 2023-01-16 1 1 病假
人员A 2023-01-20 1 1 节假日
人员A 2023-01-21 0 1 NULL
人员A 2023-01-22 0 1 NULL
人员A 2023-01-23 1 1 节假日
人员B 2023-01-01 0 1 NULL
人员B 2023-01-02 1 1 病假
人员B 2023-01-03 1 1 病假

我想要的结果如下:

姓名 请假 请假开始日期 请假结束日期 工作日 周日日
人员A 病假 2023-01-13 2023-01-16 2 4
人员A 节假日 2023-01-20 2023-01-23 2 4
人员B 病假 2023-01-02 2023-01-03 2 2
英文:

I'm trying to solve a problem in the following (simplified) dataset:

Name Date Workday Calenderday Leave
PersonA 2023-01-01 0 1 NULL
PersonA 2023-01-07 0 1 NULL
PersonA 2023-01-08 0 1 NULL
PersonA 2023-01-13 1 1 Sick
PersonA 2023-01-14 0 1 NULL
PersonA 2023-01-15 0 1 NULL
PersonA 2023-01-16 1 1 Sick
PersonA 2023-01-20 1 1 Holiday
PersonA 2023-01-21 0 1 NULL
PersonA 2023-01-22 0 1 NULL
PersonA 2023-01-23 1 1 Holiday
PersonB 2023-01-01 0 1 NULL
PersonB 2023-01-02 1 1 Sick
PersonB 2023-01-03 1 1 Sick

Where the lines with NULL in [Leave] is weekend.

What I want is a result looking like this:

Name Leave PeriodStartDate PeriodEndDate Workdays Weekdays
PersonA Sick 2023-01-13 2023-01-16 2 4
PersonA Holiday 2023-01-20 2023-01-23 2 4
PersonB Sick 2023-01-02 2023-01-03 2 2

where the difference between [Workdays] and [Weekdays] is that weekdays also counts the weekend.

What I have been trying is to first make a row (in two different ways)

ROW_NUMBER() OVER (PARTITION BY \[Name\] ORDER BY \[Date\]) as RowNo1
ROW_NUMBER() OVER (PARTITION BY \[Name\], \[Leave\] ORDER BY \[Date\]) as RowNo2

and after that to make a period base date:

DATEADD(DAY, 0 - \[RowNo1\], Date) as PeriodBaseDate1
,DATEADD(DAY, 0 - \[RowNo2\], \[Date\]) as PeriodBaseDate2

and after that do something like this:

MIN(\[Date\]) as PeriodStartDate
,MAX(\[Dato\]) as PeriodEndDate
,SUM(\[Calenderday\]) as Weekdays
,SUM(\[Workday\]) as Workdays
GROUP BY \[PeriodBaseDate (1 or 2?)\], \[Leave\], \[Name\]

But whatever I do I can't seem to get it to count the weekends in the periods.

It doesn't have to include my try with the RowNo, PeriodBaseDate etc.

答案1

得分: 1

以下是翻译好的部分:

我不确定你尝试做什么。根据我的理解,下面的脚本会产生预期的输出。

SELECT Name, Leave, Min(Date) PeriodStartDate,Max(Date) PeriodEndDate, SUM(Workday) Workdays, DATEDIFF(DAY,Min(Date),Max(Date))+ 1  Weekdays from YourTable
WHERE Leave IS NOT NULL
GROUP BY Name, Leave
英文:

I am not sure what you are trying to do. Based on what I understood, below script gives the expected output.

SELECT Name, Leave, Min(Date) PeriodStartDate,Max(Date) PeriodEndDate, SUM(Workday) Workdays, DATEDIFF(DAY,Min(Date),Max(Date))+ 1  Weekdays from YourTable
WHERE Leave IS NOT NULL
GROUP BY Name, Leave

答案2

得分: 1

由于我们没有您的实际完整解决方案,我提供了一个完整的可工作的解决方案。首先,我使用LAST_VALUE来为所有行提供一个Leave的值(前提是之前存在非NULL的值)。

一旦我这样做,您就会面临一个间隙和孤立问题,可以根据它进行聚合。

我假设您正在使用SQL Server 2022,这是写作时的最新版本,因为未提供版本详细信息,因此可以使用IGNORE NULLS语法。

SELECT *
INTO dbo.YourTable
FROM ('PersonA',CONVERT(date,'2023-01-01'),0,1,NULL),
      ('PersonA',CONVERT(date,'2023-01-07'),0,1,NULL),
      ('PersonA',CONVERT(date,'2023-01-08'),0,1,NULL),
      ('PersonA',CONVERT(date,'2023-01-13'),1,1,'Sick'),
      ('PersonA',CONVERT(date,'2023-01-14'),0,1,NULL),
      ('PersonA',CONVERT(date,'2023-01-15'),0,1,NULL),
      ('PersonA',CONVERT(date,'2023-01-16'),1,1,'Sick'),
      ('PersonA',CONVERT(date,'2023-01-20'),1,1,'Holiday'),
      ('PersonA',CONVERT(date,'2023-01-21'),0,1,NULL),
      ('PersonA',CONVERT(date,'2023-01-22'),0,1,NULL),
      ('PersonA',CONVERT(date,'2023-01-23'),1,1,'Holiday'),
      ('PersonB',CONVERT(date,'2023-01-01'),0,1,NULL),
      ('PersonB',CONVERT(date,'2023-01-02'),1,1,'Sick'),
      ('PersonB',CONVERT(date,'2023-01-03'),1,1,'Sick'))V(Name,Date,Workday,Calenderday,Leave);
GO
WITH Leaves AS(
    SELECT Name,
           [Date],
           Workday,
           Calenderday, --It's spelt Calendar, you should correct this typographical error as objects with typos lead to further problems.
           --Leave,
           LAST_VALUE(Leave) IGNORE NULLS OVER (PARTITION BY Name ORDER BY Date
                                                ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Leave
    FROM dbo.YourTable YT),
LeaveGroups AS(
    SELECT Name,
           [Date],
           Workday,
           CalenderDay,
           Leave,
           ROW_NUMBER() OVER (PARTITION BY Name ORDER BY Date) - 
           ROW_NUMBER() OVER (PARTITION BY Name, Leave ORDER BY Date) AS Grp
    FROM Leaves)
SELECT Name,
       Leave,
       MIN([Date]) AS PeriodStartDate,
       MAX([Date]) AS PeriodEndDate,
       SUM(WorkDay) AS WorkDays, --Assumes Workday is not a bit, if it is, CAST or CONVERT it to an int
       DATEDIFF(DAY,MIN([Date]), MAX([Date]))+1 AS Weekdays
       --SUM(CASE WHEN (DATEPART(WEEKDAY,[Date])  + @@DATEFIRST + 5) % 7 BETWEEN 0 AND 4 THEN 1 END) AS Weekdays --This method is language agnostic
FROM LeaveGroups
WHERE Leave IS NOT NULL
GROUP BY Name,
         Leave,
         Grp
ORDER BY Name,
         PeriodStartDate;

GO
DROP TABLE dbo.YourTable;
英文:

As we don't have your actual full solutions, I've provided a full working one. I firstly use LAST_VALUE to have all the rows have a value for their Leave (provided there was a non-NULL value previously).

Once I do that, you have a gaps and island problem, and can aggregate based on that.

I assume you are using SQL Server 2022, the latest version of SQL Server at the time of writing, as no version details are given and thus have access to the IGNORE NULLS syntax.

SELECT *
INTO dbo.YourTable
FROM (VALUES('PersonA',CONVERT(date,'2023-01-01'),0,1,NULL),
            ('PersonA',CONVERT(date,'2023-01-07'),0,1,NULL),
            ('PersonA',CONVERT(date,'2023-01-08'),0,1,NULL),
            ('PersonA',CONVERT(date,'2023-01-13'),1,1,'Sick'),
            ('PersonA',CONVERT(date,'2023-01-14'),0,1,NULL),
            ('PersonA',CONVERT(date,'2023-01-15'),0,1,NULL),
            ('PersonA',CONVERT(date,'2023-01-16'),1,1,'Sick'),
            ('PersonA',CONVERT(date,'2023-01-20'),1,1,'Holiday'),
            ('PersonA',CONVERT(date,'2023-01-21'),0,1,NULL),
            ('PersonA',CONVERT(date,'2023-01-22'),0,1,NULL),
            ('PersonA',CONVERT(date,'2023-01-23'),1,1,'Holiday'),
            ('PersonB',CONVERT(date,'2023-01-01'),0,1,NULL),
            ('PersonB',CONVERT(date,'2023-01-02'),1,1,'Sick'),
            ('PersonB',CONVERT(date,'2023-01-03'),1,1,'Sick'))V(Name,Date,Workday,Calenderday,Leave);
GO
WITH Leaves AS(
    SELECT Name,
           [Date],
           Workday,
           Calenderday, --It's spelt Calendar, you should correct this typopgraphical error as objects with typoes lead to further problems.
           --Leave,
           LAST_VALUE(Leave) IGNORE NULLS OVER (PARTITION BY Name ORDER BY Date
                                                ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Leave
    FROM dbo.YourTable YT),
LeaveGroups AS(
    SELECT Name,
           [Date],
           Workday,
           CalenderDay,
           Leave,
           ROW_NUMBER() OVER (PARTITION BY Name ORDER BY Date) - 
           ROW_NUMBER() OVER (PARTITION BY Name, Leave ORDER BY Date) AS Grp
    FROM Leaves)
SELECT Name,
       Leave,
       MIN([Date]) AS PeriodStartDate,
       MAX([Date]) AS PeriodEndDate,
       SUM(WorkDay) AS WorkDays, --Assumes Workday is not a bit, if it is, CAST or CONVERT it to a int
       DATEDIFF(DAY,MIN([Date]), MAX([Date]))+1 AS Weekdays
       --SUM(CASE WHEN (DATEPART(WEEKDAY,[Date])  + @@DATEFIRST + 5) % 7 BETWEEN 0 AND 4 THEN 1 END) AS Weekdays --This method is language agnostic
FROM LeaveGroups
WHERE Leave IS NOT NULL
GROUP BY Name,
         Leave,
         Grp
ORDER BY Name,
         PeriodStartDate;

GO
DROP TABLE dbo.YourTable;

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  • 本文由 发表于 2023年2月6日 18:04:43
  • 转载请务必保留本文链接:https://go.coder-hub.com/75359849.html
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