I want to create a query that breakdowns a date period into 10 days sub-periods
So a period of 2022-04-15 to 2022-05-01 should be broken into
2022-04-15 2022-04-24
2022-04-25 2022-05-01
The period could be one day (2022-04-15 to 2022-04-15) or even years
Any help appreciated
Thank you
A Tally would be a much more performant approach:
DECLARE #Start date = '20220415',
#End date = '20220501',
#Days int = 10;
WITH N AS (
SELECT N
FROM (VALUES(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL))N(N)),
Tally AS(
SELECT 0 AS I
UNION ALL
SELECT TOP (DATEDIFF(DAY,#Start,#End)/#Days)
ROW_NUMBER() OVER (ORDER BY (SELECT NULL))
FROM N N1, N N2, N N3, N N4) --Up to 1,000 rows. Add more cross joins for more rows
SELECT DATEADD(DAY, T.I*#Days,#Start),
CASE WHEN DATEADD(DAY, ((T.I+1)*#Days)-1,#Start) > #End THEN #END ELSE DATEADD(DAY, ((T.I+1)*#Days)-1,#Start) END
FROM Tally T;
You can use a recursive cte. A rough outline is as follows:
create table #test (
id int identity primary key,
date1 date,
date2 date
);
insert into #test (date1, date2) values
('2022-04-15', '2022-05-01'),
('2022-04-15', '2022-04-15');
with rcte as (
select id, date1 as date1_, dateadd(day, 10, date1) as date2_, date2 as enddate
from #test
union all
select id, date2_, dateadd(day, 10, date2_), enddate
from rcte
where date2_ <= enddate
)
select id, date1_, dateadd(day, -1, date2_)
from rcte
order by 1, 2
DB<>Fiddle
Does this help?
declare #fromdate date=cast('2022-04-15' as date);
declare #todate date=cast('2022-05-01' as date);
WITH cte_dates(tendays)
AS (
SELECT
#fromdate
UNION ALL
SELECT
case when dateadd(d,10,tendays) > #todate then #todate else dateadd(d,10,tendays) end
FROM
cte_dates
WHERE tendays < dateadd(d,-9,#todate)
)
SELECT
tendays,case when dateadd(d,9,tendays) > #todate then #todate else dateadd(d,9,tendays) end
FROM
cte_dates;
DB<>Fiddle
Related
I need to add dates from 01-01-2011 to 31-12-2014 in format: dd-mm-yyyy, how can I do this? I mean something like this:
SET #Date = '01/01/2011';
WHILE #Date <'31/12/2014'
BEGIN
INSERT INTO Calendar(DataKal) VALUES (#Date);
SET #Date = #Date + 1;
END
I am using SQL Server 2014.
If what you are saying is you want to INSERT a row for each date between 2 dates then the best, and by far fastest, method to do this is with a Tally:
DECLARE #StartDate date = '20110101',
#EndDate date = '20141230'; --Seems odd you want to omit 31 December 2014
WITH N AS(
SELECT N
FROM (VALUES(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL))N(N)),
Tally AS(
SELECT TOP (DATEDIFF(DAY, #StartDate, #EndDate) + 1)
ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) - 1 AS I
FROM N N1, N N2, N N3, N N4) --10,000 rows
INSERT INTO dbo.Calendar (DataKal)
SELECT DATEADD(DAY, T.I, #StartDate)
FROM Tally T;
You can use a recursive CTE to define the dates and load them:
with cte as (
select convert(date, '2011-01-01') as dte
union all
select dateadd(day, 1, dte)
from cte
where dte < '2014-12-31'
)
insert into calendar (datakal)
select dte
from cte
option (maxrecursion 0);
Note that this inserts the date using the internal format. If you want to see it in a particular format -- such as dd-mm-yyyy -- then you can add a second column. I would suggest adding the column as a computed column:
alter table calendar add dd_mm_yyyy as (convert(varchar(10), datakal, 105));
Here is a db<>fiddle.
I want to find the total number of days in a date range that overlap a table of date ranges.
For example, I have 7 days between 2 dates in the table below. I want to find the days between between them that also fall this date range: 2019-08-01 to 2019-08-30.
It should return 1 day.
This is the data source query:
SELECT LeaveId, UserId, StartDate, EndDate, Days
FROM TblLeaveRequest
WHERE UserId = 218
LeaveID UserID StartDate EndDate Days
----------- ----------- ----------------------- ----------------------- -----------
22484 218 2019-07-26 00:00:00.000 2019-08-01 00:00:00.000 7
I believe this might help you:
--create the table
SELECT
22484 LeaveID,
218 UserID,
CONVERT(DATETIME,'7/26/2019') StartDate,
CONVERT(DATETIME,'8/1/2019') EndDate,
7 Days
INTO #TblLeaveRequest
--Range Paramters
DECLARE #StartRange AS DATETIME = '8/1/2019'
DECLARE #EndRange AS DATETIME = '8/30/2019'
--Find sum of days between StartDate and EndDate that intersect the range paramters
--for UserId=218
SELECT
SUM(
DATEDIFF(
DAY
,CASE WHEN #StartRange < StartDate THEN StartDate ELSE #StartRange END
,DATEADD(DAY, 1, CASE WHEN #EndRange > EndDate THEN EndDate ELSE #EndRange END)
)
) TotalDays
from #TblLeaveRequest
where UserId=218
This assumes that no start dates are greater than end dates. It also assumes that the range parameters always intersect some portion of the range in the table.
If the parameter ranges might not intersect then you'll have to eliminate those cases by excluding negative days:
SELECT SUM( CASE WHEN Days < 0 THEN 0 ELSE Days END ) TotalDays
FROM
(
SELECT
DATEDIFF(
DAY
,CASE WHEN #StartRange < StartDate THEN StartDate ELSE #StartRange END
,DATEADD(DAY, 1, CASE WHEN #EndRange > EndDate THEN EndDate ELSE #EndRange END)
) Days
from #TblLeaveRequest
where UserId=218
) TotalDays
if I understand your problem correctly, you have two ranges of dates and you are looking for the number of days in the intersection.
Considering a Gant chart:
Start_1 .................... End_1
Start_2 .......................End_2
If you can create a table structure like
LeaveID UserID StartDate_1 EndDate_1 StartDate_2 EndDate_2
----------- ----------- ---------- --------- ---------- ---------
22484 218 2019-07-26 2019-08-01 2019-08-01 2019-08-30
You can determine the number of days by
select
leaveID
, UserID
,case
when StartDate_2 <= EndDate_1 then datediff(day,StartDate_2,EndDate_1) + 1
else 0
end as delta_days_intersection
from table
I hope that helps
You can use a numbers (Tally) table to count the number of days:
SQL Fiddle
MS SQL Server 2017 Schema Setup:
CREATE TABLE LeaveRequest
(
LeaveId Int,
UserId Int,
StartDate Date,
EndDate Date,
Days Int
)
Insert Into LeaveRequest
VALUES
(22484, 218, '2019-07-26','2019-08-01', 7)
Query 1:
DECLARE #StartDate Date = '2019-08-01'
DECLARE #EndDate Date = '2019-08-30'
;WITH Tally
AS
(
SELECT ROW_NUMBER() OVER (ORdER By Numbers.Num) AS Num
FROM
(
Values(1),(2),(3),(4),(5),(6),(7),(8),(9)
)Numbers(Num)
Cross APPLY
(
Values(1),(2),(3),(4),(5),(6),(7),(8),(9)
)Numbers2(Num2)
)
SELECT COUNT(DATEADD(d, Num -1, StartDate)) As NumberOfDays
FROM LeaveRequest
CROSS APPLY Tally
WHERE DATEADD(d, Num -1, StartDate) <= EndDate AND
DATEADD(d, Num -1, StartDate) >= #StartDate AND
DATEADD(d, Num -1, StartDate) <= #EndDate
Results:
| NumberOfDays |
|--------------|
| 1 |
CREATE TABLE #LeaveRequest
(
LeaveId Int,
UserId Int,
StartDate Date,
EndDate Date,
Days Int
)
Insert Into #LeaveRequest
VALUES
(22484, 218, '2019-07-26','2019-08-01', 7)
Declare #FromDate datetime='2019-08-01'
declare #ToDate datetime='2019-08-30'
;With CTE as
(
select top (DATEDIFF(day,#FromDate,#ToDate)+1)
DATEADD(day, ROW_NUMBER()over(order by (select null))-1,#FromDate) DT
from sys.objects
)
select * from #LeaveRequest LR
cross apply(select count(*)IntersectingDays
from CTE c where dt between lr.StartDate and lr.EndDate)ca
--or
--select lr.*,c.*
from CTE c
--cross apply
(select lr.* from #LeaveRequest LR
where c.DT between lr.StartDate and lr.EndDate)lr
drop table #LeaveRequest
Better create one Calendar table which will always help you in other queries also.
create table CalendarDate(Dates DateTime primary key)
insert into CalendarDate WITH (TABLOCK) (Dates)
select top (1000000)
DATEADD(day, ROW_NUMBER()over(order by (select null))-1,'1970-01-01') DT
from sys.objects
Then inside CTE write this,
select top (DATEDIFF(day,#FromDate,#ToDate)+1) Dates from CalendarDate
where Dates between #FromDate and #ToDate
DECLARE #FromDate datetime = '2019-08-01'
DECLARE #ToDate datetime = '2019-08-30'
SELECT
IIF(#FromDate <= EndDate AND #ToDate >= StartDate,
DATEDIFF(day,
IIF(StartDate > #FromDate, StartDate, #FromDate),
IIF(EndDate < #ToDate, EndDate, #ToDate)
),
0) AS overlapping_days
FROM TblLeaveRequest;
I have start date, end date and name of days. How can fetch all dates between those two dates of that specific days in sql?
example data:
start_date:4/11/2018
end_date: 5/11/2018
days: monday, thursday
expected output: all dates between start and end date which comes on monday and thursday and store them in table
updated
my present code(not working)
; WITH CTE(dt)
AS
(
SELECT #P_FROM_DATE
UNION ALL
SELECT DATEADD(dw, 1, dt) FROM CTE
WHERE dt < #P_TO_DATE
)
INSERT INTO Table_name
(
ID
,DATE_TIME
,STATUS
,CREATED_DATE
,CREATED_BY
)
SELECT #P_ID
,(SELECT dt FROM CTE WHERE DATENAME(dw, dt) In ('tuesday','friday',null))
,'NOT SENT'
,CAST(GETDATE() AS DATE)
,#USER_ID
Another approach for generating dates between ranges can be like following query. This will be faster compared to CTE or WHILE loop.
DECLARE #StartDate DATETIME = '2018-04-11'
DECLARE #EndDate DATETIME = '2018-05-15'
SELECT #StartDate + RN AS DATE FROM
(
SELECT (ROW_NUMBER() OVER (ORDER BY (SELECT NULL)))-1 RN
FROM master..[spt_values] T1
) T
WHERE RN <= DATEDIFF(DAY,#StartDate,#EndDate)
AND DATENAME(dw,#StartDate + RN) IN('Monday','Thursday')
Note:
If the row count present in master..[spt_values] is not sufficient for the provided range, you can make a cross join with the same to get a bigger range like following.
SELECT (ROW_NUMBER() OVER (ORDER BY (SELECT NULL)))-1 RN
FROM master..[spt_values] T1
CROSS JOIN master..[spt_values] T2
By this you will be able to generate date between a range with gap of 6436369 days.
You can use a recursive common table expression (CTE) to generate a list of days. With datepart(dw, ...) you can filter for specific days of the week.
An example that creates a list of Mondays and Thursdays between March 1st and today:
create table ListOfDates (dt date);
with cte as
(
select cast('2018-03-01' as date) as dt -- First day of interval
union all
select dateadd(day, 1, dt)
from cte
where dt < getdate() -- Last day of interval
)
insert into ListOfDates
(dt)
select dt
from cte
where datepart(dw, dt) in (2, 5) -- 2=Monday and 5=Thursday
option (maxrecursion 0)
See it working at SQL Fiddle.
This will work for you:
DECLARE #table TABLE(
ID INT IDENTITY(1,1),
Date DATETIME,
Day VARCHAR(50)
)
DECLARE #Days TABLE(
ID INT IDENTITY(1,1),
Day VARCHAR(50)
)
INSERT INTO #Days VALUES ('Monday')
INSERT INTO #Days VALUES ('Thursday')
DECLARE #StartDate DATETIME='2018-01-01';
DECLARE #EndDate DATETIME=GETDATE();
DECLARE #Day VARCHAR(50)='Friday';
DECLARE #TempDate DATETIME=#StartDate;
WHILE CAST(#TempDate AS DATE)<=CAST(#EndDate AS DATE)
BEGIN
IF EXISTS (SELECT 1 FROM #Days WHERE DAY IN (DATENAME(dw,#TempDate)))
BEGIN
INSERT INTO #table
VALUES (
#TempDate, -- Date - datetime
DATENAME(dw,#TempDate) -- Day - varchar(50)
)
END
SET #TempDate=DATEADD(DAY,1,#TempDate)
END
SELECT * FROM #table
INSERT INTO TargetTab(dateCOL)
SELECT dateCOL
FROM tab
WHERE dateCOL >= startdate AND dateCOL <= enddate
AND (DATENAME(dw,dateCOL) ='Thursday' OR DATENAME(dw,dateCOL) = 'Monday')
Try this query to get your result.
Use a recursive CTE to generate your dates, then filter by week day.
SET DATEFIRST 1 -- 1: Monday, 7 Sunday
DECLARE #StartDate DATE = '2018-04-11'
DECLARE #EndDate DATE = '2018-05-15'
DECLARE #WeekDays TABLE (WeekDayNumber INT)
INSERT INTO #WeekDays (
WeekDayNumber)
VALUES
(1), -- Monday
(4) -- Thursday
;WITH GeneratingDates AS
(
SELECT
GeneratedDate = #StartDate,
WeekDay = DATEPART(WEEKDAY, #StartDate)
UNION ALL
SELECT
GeneratedDate = DATEADD(DAY, 1, G.GeneratedDate),
WeekDay = DATEPART(WEEKDAY, DATEADD(DAY, 1, G.GeneratedDate))
FROM
GeneratingDates AS G -- Notice that we are referencing a CTE that we are also declaring
WHERE
G.GeneratedDate < #EndDate
)
SELECT
G.GeneratedDate
FROM
GeneratingDates AS G
INNER JOIN #WeekDays AS W ON G.WeekDay = W.WeekDayNumber
OPTION
(MAXRECURSION 30000)
Try this:
declare #start date = '04-11-2018'
declare #end date = '05-11-2018'
declare #P_ID int = 1
declare #USER_ID int = 11
;with cte as(
select #start [date]
union all
select dateadd(DAY, 1, [date]) from cte
where [date] < #end
)
--if MY_TABLE doesn't exist
select #P_ID,
[date],
'NOT SENT',
cast(getdate() as date),
#USER_ID
into MY_TABLE
from cte
--here you can specify days: 1 - Sunday, 2 - Monday, etc.
where DATEPART(dw,[date]) in (2, 5)
option (maxrecursion 0)
--if MY_TABLE does exist
--insert into MY_TABLE
--select #P_ID,
-- [date],
-- 'NOT SENT',
-- cast(getdate() as date),
-- #USER_ID
--from cte
--where DATEPART(dw,[date]) in (2, 5)
--option (maxrecursion 0)
I have some dates I want to calculate which is currently done over several subqueries. Each subsequent subquery uses the result (a date) of the previous query in its calculation. E.g.
DECLARE #Date DATE = '20170101'
SELECT #foo1 = (SELECT TOP 1 dbo.DateFunction(DateField)
FROM [DateTable]
WHERE DateField <= #Date
ORDER BY DateField DESC)
SELECT #foo2 = (SELECT TOP 1 dbo.DateFunction(DateField)
FROM [DateTable]
WHERE DateField <= #foo1
ORDER BY DateField DESC)
....
SELECT #fooN = (SELECT TOP 1 dbo.DateFunction(DateField)
FROM [DateTable]
WHERE DateField <= #fooNMinus1
ORDER BY DateField DESC)
Is it possible (perhaps using CTE) to make a recursive query to achieve this for a specified number of times?
Weeks are almost always 7 days, so you can get the first one and then just add seven days. If so:
WITH dates as (
SELECT MAX(dbo.DateFunction(DateField)) as dte, 1 as counter
FROM [DateTable]
WHERE DateField <= #Date
UNION ALL
SELECT DATEADD(DAY, 7, dte), counter + 1
FROM dates
WHERE counter < #n
)
SELECT dte
FROM dates;
You can use small tally table as below
Declare #d1 date = '2017-01-01'
Declare #d2 date = '2017-12-31'
select top (datediff(day, #d1, #d2)+1) dt = DateAdd(day, Row_Number() over (order by (Select NULL))-1, #d1)
from master..spt_values s1, master..spt_values s2
Or custom tally tables
;with num as
( select * from (values (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) v(n) )
, n1 as (select n1.* from num n1, num n2, num n3, num n4) --numbers generation
select top (datediff(day, #d1, #d2)+1) dt = DateAdd(day, Row_Number() over (order by (Select NULL))-1, #d1)
from n1
Yes, you can use a recursive query. Since top and aggregates are not allowed in the recursive part, you can use the row_number() function instead.
Declare #date date = cast(getdate() as date), #n int = 10
declare #DateTable table (DateField date)
insert into #DateTable values ('2017-05-01'),('2017-05-02'),('2017-05-03'),('2017-05-04'),('2017-05-05'),('2017-05-06'),('2017-05-07'),('2017-05-08'),('2017-05-09'),('2017-05-10'),
('2017-05-11'),('2017-05-12'),('2017-05-13'),('2017-05-14'),('2017-05-15'),('2017-05-16'),('2017-05-17'),('2017-05-18'),('2017-05-19'),('2017-05-20')
;with date_rte as (
select top 1 dbo.DateFunction(DateField) datefield, 0 recursions, cast(1 as bigint) rn
from #dateTable
where datefield <= #date
order by datefield desc
union all
select dbo.DateFunction(DateField), recursions+1, ROW_NUMBER() over (order by d.datefield desc)
from #datetable d
join date_rte r on d.DateField <= r.datefield
where recursions < #n and rn = 1
)
select datefield
from date_rte
where rn=1 and recursions = #n
create table #temp (date date)
declare #X date
set #X = '2016-7-01'
declare #Y date
set #Y = cast (getdate() as date)
while(#X<=#Y)
begin
if (datename(WEEKDAY,#X) = 'Sunday')
insert into #temp values (#X)
set #X = cast(((cast(#X as datetime))+1)as date)
continue
end
select * from #temp
drop table #temp
Is that possible to write the above query using CTE recursion?
You can use a CTE to create a numbers table. You can then use the numbers table to get your dates like so:
Declare #Startdate Datetime = '2016-07-01'
Declare #EndDate Datetime = '2016-08-29'
;with
N0 as (SELECT 1 as n UNION ALL SELECT 1)
,N1 as (SELECT 1 as n FROM N0 t1, N0 t2)
,N2 as (SELECT 1 as n FROM N1 t1, N1 t2)
,N3 as (SELECT 1 as n FROM N2 t1, N2 t2)
,N4 as (SELECT 1 as n FROM N3 t1, N3 t2)
,nums as (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT 1)) as num FROM N4)
SELECT DATEADD(day,num-1,#startdate) as thedate
FROM nums
WHERE num <= DATEDIFF(day,#startdate,#enddate) + 1
and datename(WEEKDAY, DATEADD(day,num-1,#startdate)) = 'Sunday'
Each table (N0 to nums) effectively multiplies the number of rows in the previous 'table', so you end up with 65,536 rows of numbers in nums (you can do less or more by adding or removing table NX as required). Then, use the numbers table to add days to your start date(SELECT DATEADD(day,num-1,#startdate) as thedate) , where the dates returned are in your date range, and the weekday is Sunday.
Also, because the numbers in nums start at 1, we use nums-1 in our select, so as to avoid skipping over the first date in our series, effectively giving us DATEADD(day, 0, #startdate) in our first row.
You can try something like explained here: http://blog.sqlauthority.com/2009/12/29/sql-server-get-date-of-all-weekdays-or-weekends-of-the-year/
DECLARE #StartDate DATETIME
DECALRE #EndDate DATETIME
SET #StartDate = '2016-07-01'
SET #EndDate = GETDATE()
;WITH cte AS (
SELECT
1 AS DayID,
#StartDate AS FromDate,
DATENAME(dw, #StartDate) AS Dayname
UNION ALL
SELECT
cte.DayID + 1 AS DayID,
DATEADD(d, 1, cte.FromDate),
DATENAME(dw, DATEADD(d, 1, cte.FromDate)) AS Dayname
FROM cte
WHERE DATEADD(d, 1, cte.FromDate) < #EndDate
)
SELECT FromDate AS Date, Dayname
FROM cte
WHERE Dayname IN ('Sunday')
I'm thinking there's be a more efficient way to do this by using a numbers table and a factor of 7 (depending on how many dates you need to get, possible pre-calc the first sunday from your start date, then join all numbers that are factors of 7 from a numbers table), but the above works well enough also.
;with cte
as
(
select getdate() as datee
union all
select dateadd(day,1,datee)
from cte
where datediff(day,getdate(),datee)<100
)
select * from cte
where datename(WEEKDAY,datee) = 'Sunday'
You will hit max recursion limit as well ,to avoid..that use something like below..
option ( MaxRecursion 0 )
I would solve this using numbers table if i am not constrained by the need to use recursive cte ,which is also way faster than Recursive cte..
select
dateadd(day,n,getdate()) as datee
from numbers
where n<100 and datename(weekday,dateadd(day,n,getdate()))='sunday'
To learn why you need numbers table,check this link..https://dba.stackexchange.com/questions/11506/why-are-numbers-tables-invaluable