I have a report we will call ReportOne, in this ReportOne I am querying the data for this report with a stored procedure. The stored procedure query returns two values which are 'TravelDate' and 'Status'.
My report has four fields, 'BeginDate', 'EndDate', 'Status', and 'Days'.
My issue is this, I need to group the report by both the 'Status' and consecutive days. Consecutive days coming from TravelDate.
'BeginDate' will be the first new date
'EndDate' will be the last consecutive date.
'Status' will be status.
'Days' will be the number of consecutive days.
Example,
TravelDate | Status
1/1/2001 | Leave
1/2/2001 | Leave
1/3/2001 | Leave
1/5/2001 | Leave
1/6/2001 | Travel
The report will then look as follows.
BeginDate | EndDate | Status | Days
1/1/2001 | 1/3/2001 | Leave | 3
1/5/2001 | 1/5/2001 | Leave | 1
1/6/2001 | 1/6/2001 | Travel | 1
Example
Declare #YourTable Table ([TravelDate] date,[Status] varchar(50))
Insert Into #YourTable Values
('1/1/2001','Leave')
,('1/2/2001','Leave')
,('1/3/2001','Leave')
,('1/5/2001','Leave')
,('1/6/2001','Travel')
Select BeginDate=min(TravelDate)
,EndDate =max(TravelDate)
,Status =max(Status)
,Days =datediff(DAY,min(TravelDate),max(TravelDate))+1
From (
Select *
,Grp = DateDiff(DAY,'1900-01-01',TravelDate) - row_number() over (partition by status order by TravelDate)
From #YourTable
) A
Group By Grp
Order By BeginDate
Returns
BeginDate EndDate Status Days
2001-01-01 2001-01-03 Leave 3
2001-01-05 2001-01-05 Leave 1
2001-01-06 2001-01-06 Travel 1
EDIT -- Capture from Stored Procedure -- #YourTable Structure must match the Structure of Stored Procedure
Declare #YourTable Table ([TravelDate] date,[Status] varchar(50))
Insert Into #YourTable
Exec youStoredProcedure
Select BeginDate=min(TravelDate)
,EndDate =max(TravelDate)
,Status =max(Status)
,Days =datediff(DAY,min(TravelDate),max(TravelDate))+1
From (
Select *
,Grp = DateDiff(DAY,'1900-01-01',TravelDate) - row_number() over (partition by status order by TravelDate)
From #YourTable
) A
Group By Grp
Order By BeginDate
EDIT - Nested Subquery
Select BeginDate=min(TravelDate)
,EndDate =max(TravelDate)
,Status =max(Status)
,Days =datediff(DAY,min(TravelDate),max(TravelDate))+1
From (
Select *
,Grp = DateDiff(DAY,'1900-01-01',TravelDate) - row_number() over (partition by status order by TravelDate)
From (
-- Your Query Here ---
) A
) A
Group By Grp
Order By BeginDate
EDIT - Consumed from a TVF
Select BeginDate=min(TravelDate)
,EndDate =max(TravelDate)
,Status =max(Status)
,Days =datediff(DAY,min(TravelDate),max(TravelDate))+1
From (
Select *
,Grp = DateDiff(DAY,'1900-01-01',TravelDate) - row_number() over (partition by status order by TravelDate)
From [dbo].[YourTableValedFunction](Param1,Param2) src
) A
Group By Grp
Order By BeginDate
Related
I have the below kind of data and I need below kind of output.
Input:
id startdate enddate
1 21/01/2019 23/01/2019
1 23/01/2019 24/01/2019
1 24/01/2029 27/01/2019
1 29/01/2019 02/02/2019
Output:
id startdate enddate
1 21/01/2019 27/01/2019
1 29/01/2019 02/02/2019
We need to use the logic of matching the first record enddate and nth record startdate.
This is a gaps-and-islands problem, where you want to group together "adjacent" dates. Here is one approach using window functions: the idea is to compare the current start date to the end date of the "previous" row, and use a window sum to define the groups:
select id, min(startdate) startdate, max(enddate) enddate
from (
select t.*,
sum(case when startdate = lag_enddate then 0 else 1 end) over(partition by id order by startdate) grp
from (
select t.*,
lag(enddate) over(partition by id order by startdate) lag_enddate
from mytable t
) t
) t
group by id, grp
Demo on DB Fiddle - with credits to Sander for creating the DDL statements in the first place:
id | startdate | enddate
-: | :--------- | :---------
1 | 2019-01-21 | 2019-01-27
1 | 2019-01-29 | 2019-02-02
have a look at
NEXT VALUE FOR method, works 2016 and later
Use a CTE or subquery (works in 2008) where you join on your own table using the previous value as a join. Here a sample script I use showing backup growth
declare #backupType char(1)
, #DatabaseName sysname
set #DatabaseName = db_name() --> Name of current database, null for all databaseson server
set #backupType ='D' /* valid options are:
D = Database
I = Database Differential
L = Log
F = File or Filegroup
G = File Differential
P = Partial
Q = Partial Differential
*/
select backup_start_date
, backup_finish_date
, DurationSec
, database_name,backup_size
, PreviouseBackupSize
, backup_size-PreviouseBackupSize as growth
,KbSec= format(KbSec,'N2')
FROM (
select backup_start_date
, backup_finish_date
, datediff(second,backup_start_date,b.backup_finish_date) as DurationSec
, b.database_name
, b.backup_size/1024./1024. as backup_size
,case when datediff(second,backup_start_date,b.backup_finish_date) >0
then ( b.backup_size/1024.)/datediff(second,backup_start_date,b.backup_finish_date)
else 0 end as KbSec
-- , b.compressed_backup_size
, (
select top (1) p.backup_size/1024./1024.
from msdb.dbo.backupset p
where p.database_name = b.database_name
and p.database_backup_lsn< b.database_backup_lsn
and type=#backupType
order by p.database_backup_lsn desc
) as PreviouseBackupSize
from msdb.dbo.backupset as b
where #DatabaseName IS NULL OR database_name =#DatabaseName
and type=#backupType
)as A
order by backup_start_date desc
using a "cursor local fast_forward" to loop over the data on a row-by-row and use a temporary table where you store & compaire prev value
Here is a solution with common table expressions that could work.
Sample data
create table data
(
id int,
startdate date,
enddate date
);
insert into data (id, startdate, enddate) values
(1, '2019-01-21', '2019-01-23'),
(1, '2019-01-23', '2019-01-24'),
(1, '2019-01-24', '2019-01-27'),
(1, '2019-01-29', '2019-02-02');
Solution
-- determine start dates
with cte_start as
(
select s.id,
s.startdate
from data s
where not exists ( select 'x'
from data e
where e.id = s.id
and e.enddate = s.startdate )
),
-- determine date boundaries
cte_startnext as
(
select s.id,
s.startdate,
lead(s.startdate) over (partition by s.id order by s.startdate) as startdate_next
from cte_start s
)
-- determine periods
select sn.id,
sn.startdate,
e.enddate
from cte_startnext sn
cross apply ( select top 1 e.enddate
from data e
where e.id = sn.id
and e.startdate >= sn.startdate
and (e.startdate < sn.startdate_next or sn.startdate_next is null)
order by e.enddate desc ) e
order by sn.id,
sn.startdate;
Result
id startdate enddate
-- ---------- ----------
1 2019-01-21 2019-01-27
1 2019-01-29 2019-02-02
Fiddle to see build up of solution and intermediate CTE results.
I have a data ranges with start and end date for a persons, I want to get the continuous date ranges only per persons:
Input:
NAME | STARTDATE | END DATE
--------------------------------------
MIKE | **2019-05-15** | 2019-05-16
MIKE | 2019-05-17 | **2019-05-18**
MIKE | 2020-05-18 | 2020-05-19
Expected output like:
MIKE | **2019-05-15** | **2019-05-18**
MIKE | 2020-05-18 | 2020-05-19
So basically output is MIN and MAX for each continuous period for the person.
Appreciate any help.
I have tried the below query:
With N AS ( SELECT Name, StartDate, EndDate
, LastStop = MAX(EndDate)
OVER (PARTITION BY Name ORDER BY StartDate, EndDate
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING) FROM Table ), B AS ( SELECT Name, StartDate, EndDate
, Block = SUM(CASE WHEN LastStop Is Null Then 1
WHEN LastStop < StartDate Then 1
ELSE 0
END)
OVER (PARTITION BY Name ORDER BY StartDate, LastStop) FROM N ) SELECT Name
, MIN(StartDate) DateFrom
, MAX(EndDate) DateTo FROM B GROUP BY Name, Block ORDER BY Name, Block
But its not considering the continuous period. It's showing the same input.
This is a type of gap-and-islands problem. There is no need to expand the data out by day! That seems very inefficient.
Instead, determine the "islands". This is where there is no overlap -- in your case lag() is sufficient. Then a cumulative sum and aggregation:
select name, min(startdate), max(enddate)
from (select t.*,
sum(case when prev_enddate >= dateadd(day, -1, startdate) then 0 else 1 end) over
(partition by name order by startdate) as grp
from (select t.*,
lag(enddate) over (partition by name order by startdate) as prev_enddate
from t
) t
) t
group by name, grp;
Here is a db<>fiddle.
Here is an example using an ad-hoc tally table
Example or dbFiddle
;with cte as (
Select A.[Name]
,B.D
,Grp = datediff(day,'1900-01-01',D) - dense_rank() over (partition by [Name] Order by D)
From YourTable A
Cross Apply (
Select Top (DateDiff(DAY,StartDate,EndDate)+1) D=DateAdd(DAY,-1+Row_Number() Over (Order By (Select Null)),StartDate)
From master..spt_values n1,master..spt_values n2
) B
)
Select [Name]
,StartDate= min(D)
,EndDate = max(D)
From cte
Group By [Name],Grp
Returns
Name StartDate EndDate
MIKE 2019-05-15 2019-05-18
MIKE 2020-05-18 2020-05-19
Just to help with the Visualization, the CTE generates the following
This will give you the same result
SELECT subquery.name,min(subquery.startdate),max(subquery.enddate1)
FROM (SELECT NAME,startdate,
CASE WHEN EXISTS(SELECT yt1.startdate
FROM t yt1
WHERE yt1.startdate = DATEADD(day, 1, yt2.enddate)
) THEN null else yt2.enddate END as enddate1
FROM t yt2) as subquery
GROUP by NAME, CAST(MONTH(subquery.startdate) AS VARCHAR(2)) + '-' + CAST(YEAR(subquery.startdate) AS VARCHAR(4))
For the CASE WHEN EXISTS I refered to SQL CASE
For the group by month and year you can see this GROUP BY MONTH AND YEAR
DB_FIDDLE
This question already has an answer here:
SQL: Gaps and Islands, Grouped dates
(1 answer)
Closed 5 years ago.
I have the following dataset:
enter image description here
Here is script for this data:
;with dataset AS (
select 'EMP01' AS EMP_ID,CAST('2018-01-01' AS DATE) AS PERIOD_START,CAST('2018-01-31' AS DATE) AS PERIOD_END,CAST('2018-01-07' AS DATE) AS CUT_DATE
UNION
select 'EMP01' AS EMP_ID,CAST('2018-01-01' AS DATE) AS PERIOD_START,CAST('2018-01-31' AS DATE) AS PERIOD_END,CAST('2018-01-15' AS DATE) AS CUT_DATE
UNION
select 'EMP02' AS EMP_ID,CAST('2018-01-01' AS DATE) AS PERIOD_START,CAST('2018-01-31' AS DATE) AS PERIOD_END,CAST('2018-01-09' AS DATE) AS CUT_DATE
)
select *
from dataset
I need to divide these periods (PERIOD_START and PERIOD_END) by CUT_DATE (exclude cut dates from that periods) The number of cut dates could be any (3,5,8 etc).
Expecting result for the dataset above is:
If your version of SQL Server supports LAG, you can use this.
SELECT EMPLOYEE_ID,
ITEM_TYPE,
MIN(APPLY_DATE) AS STARTDATE,
MAX(APPLY_DATE) AS ENDDATE
FROM
(SELECT T.*,
SUM(CASE WHEN PREV_TYPE=ITEM_TYPE THEN 0 ELSE 1 END)
OVER(PARTITION BY EMPLOYEE_ID ORDER BY APPLY_DATE) AS GRP
FROM (SELECT D.*,
LAG(ITEM_TYPE) OVER(PARTITION BY EMPLOYEE_ID ORDER BY APPLY_DATE) AS PREV_TYPE
FROM DATA D
) T
) T
WHERE ITEM_TYPE IN ('Sickness','Vacation')
GROUP BY EMPLOYEE_ID,ITEM_TYPE,GRP
The logic is to get the previous row's item_type (based on ascending order of apply_date) and compare it with the current row's value. If they are equal, they belong to the same group. Else you start a new group. This is done in the sum window function. After groups are assigned, you just need to get the max and min date for an employee_id,item_type.
Sample Demo
You would use the LAG function.
If you order by something, the LAG function gives the previous value;
a full description can be found at: http://www.sqlservercentral.com/articles/T-SQL/106783/
Take a look at vkp's answer for a full query
This is another way if way if lag is supported.
Rextester Sample
with tbl as
(select d.*
,case when (item_type = lag(item_type) over (partition by employee_id order by apply_date))
then 0
else 1
end grp_tmp
from DATA2 d
where
item_type <> 'Worked'
)
,tbl2 as
(select t.*
,sum(grp_tmp) over (order by employee_id,apply_date
rows between unbounded preceding and current row
)
as grp
from tbl t
)
select
EMPLOYEE_ID
,ITEM_TYPE
,(CONVERT(VARCHAR(24),min(apply_date),103)
+' - '
+CONVERT(VARCHAR(24),max(apply_date),103)
) as range
from tbl2
group by EMPLOYEE_ID,
ITEM_TYPE
,grp
order by
employee_id
,min(apply_date);
Output
+-------------+-----------+-------------------------+
| EMPLOYEE_ID | ITEM_TYPE | range |
+-------------+-----------+-------------------------+
| 1 | Sickness | 23/05/2017 - 24/05/2017 |
| 1 | Vacation | 26/05/2017 - 29/05/2017 |
| 1 | Sickness | 01/06/2017 - 01/06/2017 |
| 2 | Sickness | 25/05/2017 - 30/05/2017 |
+-------------+-----------+-------------------------+
In order to understand the question I will explain the result expected.
I have a db table where I save some data of the activity of the current day. Then I want to sum some numeric fields and the last register of the text fields, using a filter between two dates.
Example:
•DB TABLE
ID|CALLS|RESULT | DATE
1 | 2 |FAIL |15/09/16
1 | 1 |ERROR |16/09/16
1 | 3 |OK |17/09/16
•SUM BETWEEN 15 and 17
ID|TOTAL CALLS|LAST RESULT
1 | 6 | OK
•SUM BETWEEN 15 and 16
ID|TOTAL CALLS|LAST RESULT
1 | 3 | ERROR
-Would this be the solution?
SELECT DISTINCT ID,
TOTAL_CALLS=SUM(CALLS),
LAST_RESULT= (
SELECT RESULT FROM TABLE T2 where T2.DATE between MIN(T1.DATE) and MAX (T1.DATE) and T1.ID=T2.ID
)
FROM TABLE T1
WHERE
TIME BETWEEN 15/09/16 and 17/09/16
GROUP BY ID
Thank you very much!
Regards
Use the below query.
;WITH cte_1
AS
(SELECT ID,SUM(CALLS)OVER( PARTITION BY ID) [TOTAL CALLS]
,Result [LAST RESULT]
,ROW_NUMBER()OVER( PARTITION BY ID ORDER BY [DATE] desc) RNO
from #YourTable T
WHERE [DATE] between '09/15/2016' AND '09/16/16')
SELECT ID,[TOTAL CALLS],[LAST RESULT]
FROM cte_1
WHERE Rno=1
You can use TOP 1 WITH TIES with ORDERING to get what you need:
DECLARE #dFrom date = '2016-09-15',
#dTo date = '2016-09-16'
SELECT TOP 1 WITH TIES
ID,
SUM(CALLS) OVER (PARTITION BY ID) [TOTAL CALLS],
RESULT [LAST RESULT]
FROM YourTable
WHERE [DATE] between #dFrom and #dTo
ORDER BY ROW_NUMBER() OVER (PARTITION BY ID ORDER BY [DATE]) DESC
You can use max date to get result column value.
DECLARE #FromDate DATETIME= '15 SEP 2016'
DECLARE #ToDate DATETIME= '16 SEP 2016'
SELECT ID , SUM(CALLS) , ( SELECT RESULT
FROM #yourTable
WHERE [DATE] = #ToDate
) RESULT
FROM **#yourTable**
WHERE [DATE] BETWEEN #FromDate AND #ToDate
GROUP BY ID
Using SQL Server 2008 R2,
I'm trying to combine date ranges into the maximum date range given that one end date is next to the following start date.
The data is about different employments. Some employees may have ended their employment and have rejoined at a later time. Those should count as two different employments (example ID 5). Some people have different types of employment, running after each other (enddate and startdate neck-to-neck), in this case it should be considered as one employment in total (example ID 30).
An employment period that has not ended has an enddate that is null.
Some examples is probably enlightening:
declare #t as table (employmentid int, startdate datetime, enddate datetime)
insert into #t values
(5, '2007-12-03', '2011-08-26'),
(5, '2013-05-02', null),
(30, '2006-10-02', '2011-01-16'),
(30, '2011-01-17', '2012-08-12'),
(30, '2012-08-13', null),
(66, '2007-09-24', null)
-- expected outcome
EmploymentId StartDate EndDate
5 2007-12-03 2011-08-26
5 2013-05-02 NULL
30 2006-10-02 NULL
66 2007-09-24 NULL
I've been trying different "islands-and-gaps" techniques but haven't been able to crack this one.
The strange bit you see with my use of the date '31211231' is just a very large date to handle your "no-end-date" scenario. I have assumed you won't really have many date ranges per employee, so I've used a simple Recursive Common Table Expression to combine the ranges.
To make it run faster, the starting anchor query keeps only those dates that will not link up to a prior range (per employee). The rest is just tree-walking the date ranges and growing the range. The final GROUP BY keeps only the largest date range built up per starting ANCHOR (employmentid, startdate) combination.
SQL Fiddle
MS SQL Server 2008 Schema Setup:
create table Tbl (
employmentid int,
startdate datetime,
enddate datetime);
insert Tbl values
(5, '2007-12-03', '2011-08-26'),
(5, '2013-05-02', null),
(30, '2006-10-02', '2011-01-16'),
(30, '2011-01-17', '2012-08-12'),
(30, '2012-08-13', null),
(66, '2007-09-24', null);
/*
-- expected outcome
EmploymentId StartDate EndDate
5 2007-12-03 2011-08-26
5 2013-05-02 NULL
30 2006-10-02 NULL
66 2007-09-24 NULL
*/
Query 1:
;with cte as (
select a.employmentid, a.startdate, a.enddate
from Tbl a
left join Tbl b on a.employmentid=b.employmentid and a.startdate-1=b.enddate
where b.employmentid is null
union all
select a.employmentid, a.startdate, b.enddate
from cte a
join Tbl b on a.employmentid=b.employmentid and b.startdate-1=a.enddate
)
select employmentid,
startdate,
nullif(max(isnull(enddate,'32121231')),'32121231') enddate
from cte
group by employmentid, startdate
order by employmentid
Results:
| EMPLOYMENTID | STARTDATE | ENDDATE |
-----------------------------------------------------------------------------------
| 5 | December, 03 2007 00:00:00+0000 | August, 26 2011 00:00:00+0000 |
| 5 | May, 02 2013 00:00:00+0000 | (null) |
| 30 | October, 02 2006 00:00:00+0000 | (null) |
| 66 | September, 24 2007 00:00:00+0000 | (null) |
SET NOCOUNT ON
DECLARE #T TABLE(ID INT,FromDate DATETIME, ToDate DATETIME)
INSERT INTO #T(ID,FromDate,ToDate)
SELECT 1,'20090801','20090803' UNION ALL
SELECT 2,'20090802','20090809' UNION ALL
SELECT 3,'20090805','20090806' UNION ALL
SELECT 4,'20090812','20090813' UNION ALL
SELECT 5,'20090811','20090812' UNION ALL
SELECT 6,'20090802','20090802'
SELECT ROW_NUMBER() OVER(ORDER BY s1.FromDate) AS ID,
s1.FromDate,
MIN(t1.ToDate) AS ToDate
FROM #T s1
INNER JOIN #T t1 ON s1.FromDate <= t1.ToDate
AND NOT EXISTS(SELECT * FROM #T t2
WHERE t1.ToDate >= t2.FromDate
AND t1.ToDate < t2.ToDate)
WHERE NOT EXISTS(SELECT * FROM #T s2
WHERE s1.FromDate > s2.FromDate
AND s1.FromDate <= s2.ToDate)
GROUP BY s1.FromDate
ORDER BY s1.FromDate
An alternative solution that uses window functions rather than recursive CTEs
SELECT
employmentid,
MIN(startdate) as startdate,
NULLIF(MAX(COALESCE(enddate,'9999-01-01')), '9999-01-01') as enddate
FROM (
SELECT
employmentid,
startdate,
enddate,
DATEADD(
DAY,
-COALESCE(
SUM(DATEDIFF(DAY, startdate, enddate)+1) OVER (PARTITION BY employmentid ORDER BY startdate ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING),
0
),
startdate
) as grp
FROM #t
) withGroup
GROUP BY employmentid, grp
ORDER BY employmentid, startdate
This works by calculating a grp value that will be the same for all consecutive rows. This is achieved by:
Determine totals days the span occupies (+1 as the dates are inclusive)
SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM #t
Cumulative sum the days spanned for each employment, ordered by startdate. This gives us the total days spanned by all the previous employment spans
We coalesce with 0 to ensure we dont have NULLs in our cumulative sum of days spanned
We do not include current row in our cumulative sum, this is because we will use the value against startdate rather than enddate (we cant use it against enddate because of the NULLs)
SELECT *, COALESCE(
SUM(daysSpanned) OVER (
PARTITION BY employmentid
ORDER BY startdate
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
)
,0
) as cumulativeDaysSpanned
FROM (
SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM #t
) inner1
Subtract the cumulative days from the startdate to get our grp. This is the crux of the solution.
If the start date increases at the same rate as the days spanned then the days are consecutive, and subtracting the two will give us the same value.
If the startdate increases faster than the days spanned then there is a gap and we will get a new grp value greater than the previous one.
Although grp is a date, the date itself is meaningless we are using just as a grouping value
SELECT *, DATEADD(DAY, -cumulativeDaysSpanned, startdate) as grp
FROM (
SELECT *, COALESCE(
SUM(daysSpanned) OVER (
PARTITION BY employmentid
ORDER BY startdate
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
)
,0
) as cumulativeDaysSpanned
FROM (
SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM #t
) inner1
) inner2
With the results
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| employmentid | startdate | enddate | daysSpanned | cumulativeDaysSpanned | grp |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 5 | 2007-12-03 00:00:00.000 | 2011-08-26 00:00:00.000 | 1363 | 0 | 2007-12-03 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 5 | 2013-05-02 00:00:00.000 | NULL | NULL | 1363 | 2009-08-08 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 30 | 2006-10-02 00:00:00.000 | 2011-01-16 00:00:00.000 | 1568 | 0 | 2006-10-02 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 30 | 2011-01-17 00:00:00.000 | 2012-08-12 00:00:00.000 | 574 | 1568 | 2006-10-02 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 30 | 2012-08-13 00:00:00.000 | NULL | NULL | 2142 | 2006-10-02 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
| 66 | 2007-09-24 00:00:00.000 | NULL | NULL | 0 | 2007-09-24 00:00:00.000 |
+--------------+-------------------------+-------------------------+-------------+-----------------------+-------------------------+
Finally we can GROUP BY grp to get the get rid of the consecutive days.
Use MIN and MAX to get the new startdate and endate
To handle the NULL enddate we give them a large value to get picked up by MAX then convert them back to NULL again
SELECT
employmentid,
MIN(startdate) as startdate,
NULLIF(MAX(COALESCE(enddate,'9999-01-01')), '9999-01-01') as enddate
FROM (
SELECT *, DATEADD(DAY, -cumulativeDaysSpanned, startdate) as grp
FROM (
SELECT *, COALESCE(
SUM(daysSpanned) OVER (
PARTITION BY employmentid
ORDER BY startdate
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING
)
,0
) as cumulativeDaysSpanned
FROM (
SELECT *, DATEDIFF(DAY, startdate, enddate)+1 as daysSpanned FROM #t
) inner1
) inner2
) inner3
GROUP BY employmentid, grp
ORDER BY employmentid, startdate
To get the desired result
+--------------+-------------------------+-------------------------+
| employmentid | startdate | enddate |
+--------------+-------------------------+-------------------------+
| 5 | 2007-12-03 00:00:00.000 | 2011-08-26 00:00:00.000 |
+--------------+-------------------------+-------------------------+
| 5 | 2013-05-02 00:00:00.000 | NULL |
+--------------+-------------------------+-------------------------+
| 30 | 2006-10-02 00:00:00.000 | NULL |
+--------------+-------------------------+-------------------------+
| 66 | 2007-09-24 00:00:00.000 | NULL |
+--------------+-------------------------+-------------------------+
We can combine the inner queries to get the query at the start of this answer. Which is shorter, but less explainable
Limitations of all this required that
there are no overlaps of startdate and enddate for an employment. This could produce collisions in our grp.
startdate is not NULL. However this could be overcome by replacing NULL start dates with small date values
Future developers can decipher the window black magic you performed
A modified script for combining all overlapping periods. For example
01.01.2001-01.01.2010
05.05.2005-05.05.2015
will give one period:
01.01.2001-05.05.2015
tbl.enddate must be completed
;WITH cte
AS(
SELECT
a.employmentid
,a.startdate
,a.enddate
from tbl a
left join tbl c on a.employmentid=c.employmentid
and a.startdate > c.startdate
and a.startdate <= dateadd(day, 1, c.enddate)
WHERE c.employmentid IS NULL
UNION all
SELECT
a.employmentid
,a.startdate
,a.enddate
from cte a
inner join tbl c on a.startdate=c.startdate
and (c.startdate = dateadd(day, 1, a.enddate) or (c.enddate > a.enddate and c.startdate <= a.enddate))
)
select distinct employmentid,
startdate,
nullif(max(enddate),'31.12.2099') enddate
from cte
group by employmentid, startdate