連續計算SQL Server天數

Question

我希望為每個人計算連續的一天咒語。

我的桌子：

CREATE TABLE Absence(
Date Date,
Code varchar(10),
Name varchar(10),
Type varchar(10)
);

INSERT INTO Absence (Date, Code, Name, Type)
VALUES ('01-10-18', 'S', 'Sam', 'Sick'),
('01-11-18','S', 'Sam', 'Sick'),
('01-12-18','S', 'Sam', 'Sick'),
('01-21-18','S', 'Sam', 'Sick'),
('01-26-18','S', 'Sam', 'Sick'),
('01-27-18','S', 'Sam', 'Sick'),
('02-12-18','S', 'Sam', 'Holiday'),
('02-13-18','S', 'Sam', 'Holiday'),
('02-18-18','S', 'Sam', 'Holiday'),
('02-25-18','S', 'Sam', 'Holiday'),
('02-10-18','S', 'Sam', 'Holiday'),
('02-13-18','F', 'Fred', 'Sick'),
('02-14-18','F', 'Fred', 'Sick'),
('02-17-18','F', 'Fred', 'Sick'),
('02-25-18','F', 'Fred', 'Sick'),
('02-28-18','F', 'Fred', 'Sick');

這是我目前擁有的代碼：

WITH CTE AS
(
SELECT 
Date,
Name, 
Type
,GroupingSet = DATEADD(DAY, ROW_NUMBER() OVER 
(PARTITION BY [Name], [Type] ORDER BY [Date]), [Date])
FROM Absence
)
SELECT 
    Name,
    StartDate = MIN(Date),
    EndDate = MAX(Date),
    Result = COUNT(Name),
    min(Type) AS [Type]
    FROM CTE

   GROUP BY Name, GroupingSet
    -- HAVING COUNT(NULLIF(Code, 0)) > 1
   ORDER BY Name, StartDate

產生結果：

| Name |  StartDate |    EndDate | Result |    Type |
|------|------------|------------|--------|---------|
| Fred | 2018-02-13 | 2018-02-13 |      1 |    Sick |
| Fred | 2018-02-14 | 2018-02-14 |      1 |    Sick |
| Fred | 2018-02-17 | 2018-02-17 |      1 |    Sick |
| Fred | 2018-02-25 | 2018-02-25 |      1 |    Sick |
| Fred | 2018-02-26 | 2018-02-28 |      1 |    Sick |
|  Sam | 2018-01-10 | 2018-01-10 |      1 |    Sick |
|  Sam | 2018-01-11 | 2018-01-11 |      1 |    Sick |
|  Sam | 2018-01-12 | 2018-01-12 |      1 |    Sick |
|  Sam | 2018-01-21 | 2018-01-21 |      1 |    Sick |
|  Sam | 2018-01-26 | 2018-01-26 |      1 |    Sick |
|  Sam | 2018-01-27 | 2018-01-27 |      1 |    Sick |
|  Sam | 2018-02-10 | 2018-02-10 |      1 | Holiday |
|  Sam | 2018-02-12 | 2018-02-12 |      1 | Holiday |
|  Sam | 2018-02-13 | 2018-02-13 |      1 | Holiday |
|  Sam | 2018-02-18 | 2018-02-18 |      1 | Holiday |
|  Sam | 2018-02-25 | 2018-02-25 |      1 | Holiday |

在我正在尋找這樣的結果集的地方：

| Name |       Date | Result  |    Type |
|------|------------|---------|---------|
| Fred | 2018-02-13 |       2 |    Sick |
|  Sam | 2018-01-27 |       2 |    Sick |
|  Sam | 2018-02-10 |       1 | Holiday |

我需要計算連續1天以上的連續天數。 然后將其作為某人擁有多少個連續咒語的總和。 例如，弗雷德在這段時間內連續2次生病。 如果某人有星期五和星期一休息，我也需要此內容，這應該算作連續咒語。

我對如何到達那里有些迷失。 任何幫助將不勝感激。

請參閱： http ://sqlfiddle.com/#!18/88612/16

Answer 1

您可以使用以下方式獲取缺勤時間：

select name, min(date), max(date), count(*) as numdays, type
from (select a.*,
             row_number() over (partition by name, type order by date) as seqnum_ct
      from absence a
     ) a
group by name, type, dateadd(day, -seqnum_ct, date);

這是一個SQL Fiddle。

您可以添加having count(*) > 1以獲取一天或更長的期限。 這似乎很有用。 我不明白最終的輸出是什么。 該描述對我而言毫無意義。

如果您希望缺勤的天數為2天或以上，則可以將其用作子查詢/ CTE：

select name, count(*), type
from (select name, min(date) as mindate, max(date) as maxdate, count(*) as numdays, type
      from (select a.*,
                   row_number() over (partition by name, type order by date) as seqnum_ct
            from absence a
           ) a
      group by name, type, dateadd(day, -seqnum_ct, date)
     ) b
where numdays > 1
group by name, type;

Answer 2

嘗試這個：

select name, min([date]) [date], count(*) [result], [type] from (
    select *, SUM(isConsecutive) over (partition by name,[type] order by [date] rows between unbounded preceding and current row) [isConsecutiiveId]
    from (
        select *, case when dateadd(day, -1, [date]) = LAG([date]) over (partition by name,[type] order by [date]) then 0 else 1 end [isConsecutive] from #Absence
    ) a
) a group by name,[type],isConsecutiiveId

它導致的缺勤期比您的預期結果長，因為在您的數據中有更多的缺勤期。 它顯然包括您的結果，但還有更多:)

Answer 3

SELECT 
s.[Name], 'Last 12 Months' as [Date], s.[Type], COUNT(s.[numdays]) AS 
[Consecutive Spells]
FROM ( select name, min(date) AS [Date], max(date) AS [Date2], 
count(*) as numdays, type
from (select a.*, row_number() over (partition by name, type order by date) 
as seqnum_ct
  from absence a
 ) a
group by name, type, dateadd(day, -seqnum_ct, date)
HAVING count(*) > 1
)S
GROUP BY s.[Name], s.[Type]

這是我一直在尋找的結果。 -http://sqlfiddle.com/#!18/472d2/23

但是，感謝您對戈登的幫助，它使我朝着正確的方向前進！