Finding most recent date based on consecutive dates - sql

Question

I have s table that lists absences(holidays) of all employees, and what we would like to find out is who is away today, and the date that they will return.
Unfortunately, absences aren't given IDs, so you can't just retrieve the max date from an absence ID if one of those dates is today.
However, absences are given an incrementing ID per day as they are inputt, so I need a query that will find the employeeID if there is an entry with today's date, then increment the AbsenceID column to find the max date on that absence.
Table Example (assuming today's date is 11/11/2014, UK format):
AbsenceID EmployeeID AbsenceDate
100 10 11/11/2014
101 10 12/11/2014
102 10 13/11/2014
103 10 14/11/2014
104 10 15/11/2014
107 21 11/11/2014
108 21 12/11/2014
120 05 11/11/2014
130 15 20/11/2014
140 10 01/03/2015
141 10 02/03/2015
142 10 03/03/2015
143 10 04/03/2015
So, from the above, we'd want the return dates to be:
EmployeeID ReturnDate
10 15/11/2014
21 12/11/2014
05 11/11/2014
Edit: note that the 140-143 range couldn't be included in the results as they appears in the future, and none of the date range of the absence are today.
Presumably I need an iterative sub-function running on each entry with today's date where the employeeID matches.

Answer 1

So based on what I believe you're asking, you want to return a list of the people that are off today and when they are expected back based on the holidays that you have recorded in the system, which should only work only on consecutive days.
SQL Fiddle Demo
Schema Setup:
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')
;
Recursive CTE to generate the output:
;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
Results:
| 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 |
Explanation:
The first SELECT in the CTE gets employees that are off today with this filter:
WHERE AbsenceDate = CAST(GETDATE() AS DATE)
This result set is then UNIONED back to the EmployeeAbsence table with a join that matches EmployeeID as well as the AbsenceDate + 1 day to find the consecutive days recursively using:
-- add a day to the cte.AbsenceDate from the first SELECT
e.AbsenceDate = DATEADD(d,1,cte.AbsenceDate)
The final SELECT simply groups the cte results by employee with the MAX AbsenceDate that has been calculated per employee.
SELECT cte.EmployeeID, MAX(cte.AbsenceDate)
FROM cte
GROUP BY cte.EmployeeID
Excluding Weekends:
I've done a quick test based on your comment and the below modification to the INNER JOIN within the CTE should exclude weekends when adding the extra days if it detects that adding a day will result in a Saturday:
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
So when you add a day: datepart(dw,DATEADD(d,1,cte.AbsenceDate)) = 7, if it results in Saturday (7), then you add 3 days instead of 1 to get Monday: DATEADD(d,3,cte.AbsenceDate).

Answer 2

You'd need to do a few things to get this data into a usable format. You need to be able to work out where a group begins and ends. This is difficult with this example because there is no straight forward grouping column.
So that we can calculate when a group starts and ends, you need to create a CTE containing all the columns and also use LAG() to get the AbsenceID and EmployeeID from the previous row for each row. In this CTE you should also use ROW_NUMBER() at the same time so that we have a way to re-order the rows into the same order again.
Something like:
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]
)
Now that we have this we can compare each row to the previous to see if the current row is in a different "group" to the previous row.
The condition would be something like:
[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
You can then use this to join the CTE to it's self to find the first row in each group with something like:
....
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)
)
))
...
You can then GROUP BY the [First].[RN] which will now act like a group id and allow you to get the start and end date of each absence group.
SELECT
[Row].[EmployeeID]
,MIN([Row].[AbsenceDate]) AS [Absence_Begin]
,MAX([Row].[AbsenceDate]) AS [Absence_End]
...
-- FROM and INNER JOIN from above
...
GROUP BY [First].[RN], [Row].[EmployeeID];
You could then put all that into a view giving you the EmployeeID with the Start and End date of each absence. You can then easily pull out the Employee's currently off with a:
WHERE CAST(CURRENT_TIMESTAMP AS date) BETWEEN [Absence_Begin] AND [Absence_End]
SQL Fiddle

Answer 3

Like another answer here, I'm going to create the leave intervals, but via a different method. First the code:
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]
By way of explanation, we first put a date value into a variable. I chose today, but this code will work for any date passed in. Next, we create a common table expression (CTE) that will add on a grouping column to your table. This is the meat of the solution, so it bears some treatment. Within a given interval, the AbsenceDate increases at a rate of one day per row. row_number() also increases at a rate of one per row. So, if we subtract a row_number() number of days from the AbsenceDate, we'll get another (arbitrary) date. The key here is to realize that that arbitrary date will be the same for every row in the interval, so we can use it to group by. From there, it's just a matter of doing just that; get the min and max per interval. Lastly, we find what intervals contain #today.