SQL function for last 12 months - sql

I am looking for a SQL-function that gives the last 12 months with Start Date and End Date. Say you pick 10.Dec, it will give a result in:
- StartDate -- EndDate
- 2013-11-01 - 2013-11-30
- 2013-10-01 - 2013-10-31
- 2013-09-01 - 2013-09-30
and so it goes for the last 12 months.
I tried modifying an old function we had, but I got totally off and confused in the end.
ALTER FUNCTION [dbo].[Last12Months](#Date date) RETURNS TABLE
AS
Return
(
with cte as (
SELECT DATEADD(mm, DATEDIFF(mm, 01, #Date), 01) AS Start,
DATEADD(mm, DATEDIFF(mm, -12, #Date), -12) AS EndDate
union all
select Start - 1, EndDate - 1 from cte
where Start >= #Date )
select CAST(Start as DATE) StartDate, CAST(EndDate as DATE) EndDate from cte)
Runned it like this:
select * from dbo.Last12Months ('2013-12-10')
and got:
- StartDate - EndDate
- 2013-12-02 - 2013-12-20
Anyone know what to do?

Please try using CTE:
ALTER FUNCTION [dbo].[Last12Months]
(
#Date datetime
) RETURNS #tbl TABLE (Start datetime, EndDate datetime)
AS
BEGIN
WITH T AS(
SELECT
DATEADD(month, DATEDIFF(month, 0, #Date), 0) AS Start,
DATEADD(d, -DAY(DATEADD(m,1,#date)),DATEADD(m,1,#date)) AS EndDate,
12 Cnt
UNION ALL
SELECT
DATEADD(month, -1, Start),
DATEADD(d, -DAY(DATEADD(m,1,Start-1)),DATEADD(m,1,Start-1)),
Cnt-1
FROM
T
WHERE
Cnt-1>0
)
INSERT INTO #tbl
(Start, EndDate)
SELECT
Start, EndDate
FROM T
RETURN
END

This seems to do the job - whether you want to put it in a function or just wherever you need to have the data:
; With Numbers as (
select ROW_NUMBER() OVER (ORDER BY number ) as n
from master..spt_values
), Months as (
select DATEADD(month,n,'20010101') as start_date,
DATEADD(month,n,'20010131') as end_date
from Numbers
)
select * from Months
where DATEDIFF(month,start_date,GETDATE()) between 0 and 11
(Substitute any other date for GETDATE() if you want to get it based on some other date)
(On my machine, this can generate any month from January 2001 on to at least the next century - it can be adjusted if you need earlier or later dates also)

Damn, you've got to be quick on SO!
Good use of CTEs: i've learnt a bit answering this...
alter function Last12Months(#d date) returns table
as
return(
with cte as (
select
dateadd(month, datepart(mm,#d)-13,
dateadd(year,datepart(yyyy,#d)-1900,0)
)
as start
union all
select dateadd(mm, 1, start) from cte
where start < #d)
select start, dateadd(mm, 1, start) ends from cte
where start < #d
)
go
select * from Last12Months('2014-06-04')
Removed conversion to varchar thanks to
Date serial in SQL?
This returns 13 months: from say June last year to this June, inclusive.
To return the previous 12 months, not including the current June, change the final start<#d to
where start < dateadd(month, datepart(mm,#d)-1,
dateadd(year,datepart(yyyy,#d)-1900,0))
The end is 00:00 hours on the first day of the next month.

check this,
Declare #i date='2013-12-10'
;with cte as
(Select dateadd(month,datediff(month,0,#i)-1,0) StartDate
,dateadd(day,-1,dateadd(month,datediff(month,0,#i),0)) EndDate ,1 rownum
Union all
select dateadd(month,-1,StartDate),dateadd(day,-1,StartDate),rownum+1 rownum from cte where rownum<12 )
select * from cte

#Lebowski Below script will give you start and end date of specified calendar months from today in chronological order
DECLARE #nMonths TINYINT
SET #nMonths = 60
SELECT FORMAT(DATEADD(month, n.n - #nMonths+1+ DATEDIFF(month, 0, GETDATE()) -1 ,0), 'yyyy-MM-dd') AS MonthStartDate
, FORMAT(DATEADD(dd, -1, DATEADD(month, n.n - #nMonths+1 + DATEDIFF(month, 0, GETDATE()),0)), 'yyyy-MM-dd') AS MonthEndDate
FROM (SELECT TOP(#nMonths) n = ROW_NUMBER() OVER (ORDER BY NAME)
FROM master.dbo.syscolumns) n
Sample output
MonthStartDate MonthEndDate
2011-04-01 2011-04-30
2011-05-01 2011-05-31
2011-06-01 2011-06-30
2011-07-01 2011-07-31
2011-08-01 2011-08-31
2011-09-01 2011-09-30
2011-10-01 2011-10-31
2011-11-01 2011-11-30
2011-12-01 2011-12-31
....

Try this it might help you
select top 12 *
from YourTable
where dateOf between #DateFrom and #DateTo
order by dateOf desc

Related

How to get date of a beginning of first work weekday even if the Monday is from last month?

My goal to is get get query that will return weekdays in a month. I can get the days of the month but I need to get dates starting from monday through Friday even if the Monday may be in the preceding month.
Example April 1st is a wednesday so I would need to bring back March 30th and 31st. And the last date returned would be by May 1st as that is the last friday that contains some April days..
If interested in a helper function, I have TVF which generates a calendar.
Example
Select * from [dbo].[tvf-Date-Calendar-Wide]('2020-04-01')
Returns
So, with a little tweak, we get can
Select WeekNr = RowNr
,B.*
From [dbo].[tvf-Date-Calendar-Wide]('2020-04-01') A
Cross Apply ( values (Mon)
,(Tue)
,(Wed)
,(Thu)
,(Fri)
) B(Date)
Which Returns
WeekNr Date
1 2020-03-30
1 2020-03-31
1 2020-04-01
1 2020-04-02
1 2020-04-03
2 2020-04-06
2 2020-04-07
2 2020-04-08
...
5 2020-04-29
5 2020-04-30
5 2020-05-01
The Function If Interested
CREATE FUNCTION [dbo].[tvf-Date-Calendar-Wide] (#Date1 Date)
Returns Table
Return (
Select RowNr,[Sun],[Mon],[Tue],[Wed],[Thu],[Fri],[Sat]
From (
Select D
,DOW=left(datename(WEEKDAY,d),3)
,RowNr = sum(Flg) over (order by D)
From (
Select D,Flg=case when datename(WEEKDAY,d)= 'Sunday' then 1 else 0 end
From (Select Top (42) D=DateAdd(DAY,-7+Row_Number() Over (Order By (Select Null)),#Date1) From master..spt_values n1 ) A
) A
) src
Pivot (max(d) for DOW in ([Sun],[Mon],[Tue],[Wed],[Thu],[Fri],[Sat]) )pvg
Where [Sun] is not null
and [Sat] is not null
)
-- Select * from [dbo].[tvf-Date-Calendar-Wide]('2020-04-01')
You first need to find the start of the week for the first day of the month, then the date for the end of the week that contains the last day of the month:
e.g.
SELECT WeekStart = DATEADD(DAY, -(DATEPART(WEEKDAY, '20200401')-1), '20200401'),
WeekEnd = DATEADD(DAY, 7-(DATEPART(WEEKDAY, '20200430')), '20200430');
Gives:
WeekStart WeekEnd
------------------------------
2020-03-29 2020-05-02
You wouldn't want to hard code the first and the last of the month, but these are fairly trivial things to get from a date:
DECLARE #Date DATE = '20200415';
SELECT MonthStart = DATEADD(MONTH, DATEDIFF(MONTH, 0, #Date), 0),
MonthEnd = EOMONTH(#Date);
Which returns
MonthStart MonthEnd
------------------------------
2020-04-01 2020-04-30
You can then just substitute this into the first query for week starts:
DECLARE #Date DATE = '20200401';
SELECT WeekStart = DATEADD(DAY, -(DATEPART(WEEKDAY, DATEADD(MONTH, DATEDIFF(MONTH, 0, #Date), 0))-1), DATEADD(MONTH, DATEDIFF(MONTH, 0, #Date), 0)),
WeekEnd = DATEADD(DAY, 7-(DATEPART(WEEKDAY, EOMONTH(#Date))), EOMONTH(#Date));
Which gives the same output as the first query with hard coded dates. This is very clunky though, so I would separate this out into a further step:
DECLARE #Date DATE = '20200401';
-- SET DATE TO THE FIRST OF THE MONTH IN CASE IT IS NOT ALREADY
SET #Date = DATEADD(MONTH, DATEDIFF(MONTH, 0, #Date), 0);
SELECT WeekStart = DATEADD(DAY, -(DATEPART(WEEKDAY, #Date)-1), #Date),
Weekend = DATEADD(DAY, 7-(DATEPART(WEEKDAY, EOMONTH(#Date))), EOMONTH(#Date));
Again, this gives the same output (2020-03-29 & 2020-05-02).
The next step is to fill in all the dates between that are weekdays. If you have a calendar table this is very simple
DECLARE #Date DATE = '20200415';
-- SET DATE TO THE FIRST OF THE MONTH IN CASE IT IS NOT ALREADY
SET #Date = DATEADD(MONTH, DATEDIFF(MONTH, 0, #Date), 0);
DECLARE #Start DATE = DATEADD(DAY, -(DATEPART(WEEKDAY, #Date)-1), #Date),
#End DATE = DATEADD(DAY, 7-(DATEPART(WEEKDAY, EOMONTH(#Date))), EOMONTH(#Date));
SELECT [Date], DayName = DATENAME(WEEKDAY, [Date])
FROM Calendar
WHERE Date >= #Start
AND Date <= #End
AND IsWeekday = 1
ORDER BY [Date];
If you don't have a calendar table, then I suggest you create one, but if you can't create one you can still generate this on the fly, by generating a set series numbers and adding these numbers to your start date:
DECLARE #Date DATE = '20200415';
-- SET DATE TO THE FIRST OF THE MONTH IN CASE IT IS NOT ALREADY
SET #Date = DATEADD(MONTH, DATEDIFF(MONTH, 0, #Date), 0);
DECLARE #Start DATE = DATEADD(DAY, -(DATEPART(WEEKDAY, #Date)-1), #Date),
#End DATE = DATEADD(DAY, 7-(DATEPART(WEEKDAY, EOMONTH(#Date))), EOMONTH(#Date));
-- GET NUMBERS FROM 0 - 50
WITH Dates (Date) AS
( SELECT TOP (DATEDIFF(DAY, #Start, #End))
DATEADD(DAY, ROW_NUMBER() OVER(ORDER BY n1.n) - 1, #Start)
FROM (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) n1 (n)
CROSS JOIN (VALUES (1),(1),(1),(1),(1)) n2 (n)
)
SELECT [Date], DayName = DATENAME(WEEKDAY, [Date])
FROM Dates
WHERE ((DATEPART(WEEKDAY, [Date]) + ##DATEFIRST) % 7) NOT IN (0, 1);
Just generate all possible dates -- up to 6 days before the month begins. Take the valid weekdays after the first Monday:
with dates as (
select dateadd(day, -6, convert(date, '2020-04-01')) as dte
union all
select dateadd(day, 1, dte)
from dates
where dte < '2020-04-30'
)
select dte
from (select d.*,
min(case when datename(weekday, dte) = 'Monday' then dte end) over () as first_monday
from dates d
) d
where datename(weekday, dte) not in ('Saturday', 'Sunday') and
dte >= first_monday;
declare #dateVal datetime = GETDATE(); --assign your date here
declare #monthFirstDate datetime = cast(YEAR(#dateVal) as varchar(4)) + '-' + DATENAME(mm, #dateVal) + '-' + cast(01 as varchar(2))
declare #monthLastDate datetime = DAteADD(day, -1, DATEADD(month, 1, cast(YEAR(#dateVal) as varchar(4)) + '-' + DATENAME(mm, #dateVal) + '-' + cast(01 as varchar(2))))
declare #startDate datetime = DATEADD(DAY, 2 - DATEPART(WEEKDAY, #monthFirstDate), CAST(#monthFirstDate AS DATE))
declare #enddate datetime = DATEADD(DAY, 6 - DATEPART(WEEKDAY, #monthLastDate), CAST(#monthLastDate AS DATE))
Select #startDate StartDate, #enddate EndDate
****Result**
--------------------------------------------------------------
StartDate | EndDate
-----------------------------|--------------------------------
2020-03-02 00:00:00.000 | 2020-04-03 00:00:00.000
-----------------------------|---------------------------------**

Start and end of each month between two dates [duplicate]

This question already has answers here:
months between two dates in sql server with starting and end date of each of them in sql server
(6 answers)
Closed 4 years ago.
Given two #START_DATE and #END_DATE parameters, I'd like to write a query that generates pairs of (month_start_date, month_end_date) for every month that exists between those two days, including the ones those dates are in.
E.g. If #START_DATE = '2018-01-14' and #END_DATE = '2018-05-04' (yyyy-MM-dd), I'd like the output to be
month_start_date, month_end_date
2018-01-01, 2018-01-31
2018-02-01, 2018-02-28
2018-03-01, 2018-03-31
2018-04-01, 2018-04-30
2018-05-01, 2018-05-31
I tend to go for recursive CTEs for this purpose:
with dates as (
select datefromparts(year(#start_date), month(#start_date), 1) as dte
union all
select dateadd(month, 1, dte)
from dates
where dateadd(month, 1, dte) <= #end_date
)
select dte as start_date, eomonth(dte) as end_date
from dates;
This works as-is for up to 100 months. For more than that, you need to use set the max recursion option.
Following query returns the required result.
declare #StartDate date = '2018-01-14'
, #EndDate date = '2018-05-04';
;with Months as (
select top (datediff(month,#StartDate,#EndDate)+1)
[month_start_date] = dateadd(month
, datediff(month, 0, #StartDate) + row_number() over (order by number) -1
, 0)
, month_end_date = dateadd(day,-1,dateadd(month
, datediff(month, 0, #StartDate) + row_number() over (order by number)
,0))
from master.dbo.spt_values
order by [month_start_date]
)
select * from Months;
You need recursive cte :
with t as (
select dateadd(day, 1, eomonth(#START_DATE, -1)) as start_dt, #END_DATE as end_dt
union all
select dateadd(mm, 1, start_dt), end_dt
from t
where dateadd(mm, 1, start_dt) < #END_DATE
)
select start_dt as month_start_date, eomonth(start_dt) as month_end_date
from t
option (maxrecursion 0);

tsql: How to retrieve the last date of each month between given date range

I have two date for example 08/08/2013 and 11/11/2013 and I need last date of each month starting from August to November in a table so that i can iterate over the table to pick those dates individually.
I know how to pick last date for any month but i am stucked with a date range.
kindly help, it will be highly appreciated.
Note : I am using Sql 2008 and date rang could be 1 month , 2 month or 6 month or a year or max too..
You can use CTE for getting all last days of the month within the defined range
Declare #Start datetime
Declare #End datetime
Select #Start = '20130808'
Select #End = '20131111'
;With CTE as
(
Select #Start as Date,Case When DatePart(mm,#Start)<>DatePart(mm,#Start+1) then 1 else 0 end as [Last]
UNION ALL
Select Date+1,Case When DatePart(mm,Date+1)<>DatePart(mm,Date+2) then 1 else 0 end from CTE
Where Date<#End
)
Select * from CTE
where [Last]=1 OPTION ( MAXRECURSION 0 )
DECLARE #tmpTable table (LastDates DATE);
DECLARE #startDate DATE = '01/01/2012'; --1 Jan 2012
DECLARE #endDate DATE = '05/31/2012'; --31 May 2012
DECLARE #tmpEndDate DATE;
SET #startDate = DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#startDate)+1,1));
SET #tmpEndDate = DATEADD(DAY, 1, #endDate);
WHILE (#startDate <= #tmpEndDate)
BEGIN
INSERT INTO #tmpTable (LastDates) values (DATEADD(DAY, -1, #startDate));
SET #startDate = DATEADD(MONTH, 1, #startDate);
END
SELECT [LastDates] FROM #tmpTable;
Output:
Example: 1
#startDate DATE = '01/01/2012'; --1 Jan 2012
#endDate DATE = '05/31/2012'; --31 May 2012
LastDates
----------
2012-01-31
2012-02-29
2012-03-31
2012-04-30
2012-05-31
Example: 2
#startDate DATE = '11/01/2011'; --1 Nov 2011
#endDate DATE = '03/13/2012'; --13 Mar 2012
LastDates
----------
2011-11-30
2011-12-31
2012-01-31
2012-02-29
I've created a table variable, filled it with all days between #startDate and #endDate and searched for max date in the month.
declare #tmpTable table (dates date)
declare #startDate date = '08/08/2013'
declare #endDate date = '11/11/2013'
while #startDate <= #endDate
begin
insert into #tmpTable (dates) values (#startDate)
set #startDate = DATEADD(DAY, 1, #startDate)
end
select max(dates) as [Last day] from #tmpTable as o
group by datepart(YEAR, dates), datepart(MONTH, dates)
Results:
Last day
2013-08-31
2013-09-30
2013-10-31
2013-11-11
To also get last day of November this can be used before loop:
set #endDate = DATEADD(day, -1, DATEADD(month, DATEDIFF(month, 0, #endDate) + 1, 0))
Following script demonstrates the script to find last day of previous, current and next month.
----Last Day of Previous Month
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE()),0))
LastDay_PreviousMonth
----Last Day of Current Month
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE())+1,0))
LastDay_CurrentMonth
----Last Day of Next Month
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE())+2,0))
LastDay_NextMonth
If you want to find last day of month of any day specified use following script.
--Last Day of Any Month and Year
DECLARE #dtDate DATETIME
SET #dtDate = '8/18/2007'
SELECT DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#dtDate)+1,0))
LastDay_AnyMonth
ResultSet:
LastDay_AnyMonth
Source - SQL Server Central.
You can use a recursive CTE to do this, note the MAXRECURSION OPTION prevents an infinite loop:
DECLARE #StartDate DATE = '2013-08-08'
DECLARE #EndDate DATE = '2013-11-11'
;WITH dateCTE
AS
(
SELECT CAST(DATEADD(M, 1,DATEADD(d, DAY(#StartDate) * -1, #StartDate)) AS DATE) EndOFMonth
UNION ALL
SELECT CAST(DATEADD(M, 2,DATEADD(d, DAY(EndOFMonth) * -1, EndOFMonth)) AS DATE)
FROM dateCTE
WHERE EndOFMonth < DATEADD(d, DAY(#EndDate) * -1, #EndDate)
)
SELECT *
FROM dateCTE
OPTION (MAXRECURSION 30);
This returns
EndOFMonth
----------
2013-08-31
2013-09-30
2013-10-31
try this
the last row(where) is optional for date filtering
declare #table table
(
thisdate date
)
insert into #table values ('12/01/2013'),('05/06/2013'),('04/29/2013'),('02/20/2013')
select *,DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,thisdate)+1,0))
LastDay from #table
where thisdate between 'givendate' and 'givendate'
The Example Below is for all dates
thisdate lastday
2013-12-01 2013-12-31 23:59:59.000
2013-05-06 2013-05-31 23:59:59.000
2013-04-29 2013-04-30 23:59:59.000
2013-02-20 2013-02-28 23:59:59.000
The following CTE gives you the last day of every month from February 1900 until the middle of the 26th century (on my machine):
;with LastDaysOfMonths as (
select DATEADD(month,
ROW_NUMBER() OVER (ORDER BY so.object_id),
'19000131') as Dt
from sys.objects so,sys.objects so1
)
select * from LastDaysOfMonths
It should be easy enough to use it as part of a larger query or to filter it down to just the dates you want. You can adjust the range of years as needed by changing the constant 19000131. The only important thing to do is make sure that you use a month that has 31 days in it and always have the constant be for day 31.
No need to use a common table expression or anything like that - this simple query will do it:
SELECT DATEADD(d, -1, DATEADD(mm, DATEDIFF(m, 0, DATEADD(m, number, '2013-08-08')) + 1, 0)) AS EndOfMonth
FROM master.dbo.spt_values
WHERE 'P' = type
AND DATEADD(m, number, '2013-08-08') < '2013-11-11';
Although the question is about the last day which #bummi has already answered.
But here is the solution for the first date which might be helpful for someone.
Get the first dates of all the months in-between the #FromDate and #ToDate.
DECLARE #FromDate DATETIME = '2019-08-13'
DECLARE #ToDate DATETIME = '2019-11-25'
;WITH CTE
AS
(
SELECT DATEADD(DAY, -(DAY(#FromDate) - 1), #FromDate) AS FirstDateOfMonth
UNION ALL
SELECT DATEADD(MONTH, 1, FirstDateOfMonth)
FROM CTE
WHERE FirstDateOfMonth < DATEADD(DAY, -(DAY(#ToDate) - 1), #ToDate)
)
SELECT * FROM CTE
Here is the result
--Result
2019-08-01 00:00:00.000
2019-09-01 00:00:00.000
2019-10-01 00:00:00.000
2019-11-01 00:00:00.000

How to get all the weekend dates of the current year in SQL?

I tried but could not get the right solution. I want an SQL query that lists all the weekend dates of the current year.
I tried this SQL query:
WITH hier(num, lvl) AS (
SELECT 0, 1
UNION ALL
SELECT 100, 1
UNION ALL
SELECT num + 1, lvl + 1
FROM hier
WHERE lvl < 100
)
SELECT lvl [Week],
convert(date,DATEADD(dw, -DATEPART(dw, DATEADD(wk,DATEDIFF(wk,0,'12/31/'+convert(nvarchar,YEAR(getdate()))), 0)+6 ),
DATEADD(wk, DATEDIFF(wk,0,'12/31/'+convert(nvarchar,YEAR(getdate()))), 0)+6 ) - num * 7,101) [End Date]
FROM hier a
where num < 52
ORDER BY [End Date] asc
Its output is like this:
Week End date
52 2012-01-14
51 2012-01-21
50 2012-01-28
49 2012-02-04
I want the dates to start from the beginning – so, the above is missing one weekend, which is 2012-07-01. Also, I want the week numbers to show as 1, 2, 3... instead of 52, 51....
Check out this blog post.
Your question is explained in detail.
DECLARE #Year AS INT,
#FirstDateOfYear DATETIME,
#LastDateOfYear DATETIME
-- You can change #year to any year you desire
SELECT #year = 2010
SELECT #FirstDateOfYear = DATEADD(yyyy, #Year - 1900, 0)
SELECT #LastDateOfYear = DATEADD(yyyy, #Year - 1900 + 1, 0)
-- Creating Query to Prepare Year Data
;WITH cte AS (
SELECT 1 AS DayID,
#FirstDateOfYear AS FromDate,
DATENAME(dw, #FirstDateOfYear) 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) < #LastDateOfYear
)
SELECT FromDate AS Date, Dayname
FROM CTE
WHERE DayName IN ('Saturday','Sunday') -- For Weekend
/*
WHERE DayName LIKE 'Sunday'
WHERE DayName NOT IN ('Saturday','Sunday') -- For Weekday
WHERE DayName LIKE 'Monday' -- For Monday
WHERE DayName LIKE 'Sunday' -- For Sunday
*/
OPTION (MaxRecursion 370)
Will this help
DECLARE #startDate DATETIME, #endDate DATETIME
SELECT #startDate = '2012-01-01', #endDate = '2012-12-31'
;WITH Calender AS (
SELECT #startDate AS dt
UNION ALL
SELECT dt + 1 FROM Calender
WHERE dt + 1 <= #endDate
)
SELECT
dt
,NameMonth = DATENAME(Month, dt)
,NameDay = DATENAME (Weekday,dt)
,WeekofYr = DATEPART(WEEK, dt) FROM Calender
WHERE DATENAME (Weekday,dt) IN ('Sunday')
Option(MaxRecursion 0)
Result(Partial)
dt NameMonth NameDay WeekofYr
2012-01-01 00:00:00.000 January Sunday 1
2012-01-08 00:00:00.000 January Sunday 2
...............................................
...............................................
2012-12-30 00:00:00.000 December Sunday 53
you can try this
DECLARE #FirstDateOfYear DATETIME
SET #FirstDateOfYear = ’2010-01-01′
SELECT DISTINCT DATEADD(d, number, #FirstDateOfYear),
CASE DATEPART(dw, DATEADD(d, number, #FirstDateOfYear))
WHEN 7 THEN ‘Saturday’
WHEN 1 THEN ‘Sunday’
ELSE ‘Work Day’
END
FROM master..spt_values
WHERE number BETWEEN 0 AND 364
AND (DATEPART(dw, DATEADD(d, number, #FirstDateOfYear)) = 1 OR DATEPART(dw, DATEADD(d, number, #FirstDateOfYear)) = 7)
ORDER BY DATEADD(d, number, #FirstDateOfYear)
Try to find the first Saturday by doing this:
Start on 2012-01-01
If it's not a Saturday, add a day
Goto 2
Then, into a temporary table, add that date and the following date (Sunday).
After that, loop the following:
Add 7 and 8 days to the last Saturday you found (you get the following Saturday and Sunday)
Check whether they are still in 2012
If they are, store them in temp table and goto 1
There may be more elegant ways, but that's my quick & dirty solution. As you didn't post any code of what you've tried, I'll leave the implementation up to you.
this also works
declare #dat datetime, #add int
set #dat = '20120101'
set #add = datepart(w,#dat)
set #add = 5 - #add -- friday
set #dat = dateadd(d,#add,#dat)
while #dat <= '20121231'
begin
print #dat
set #dat = dateadd(d,7,#dat)
end
;with AllDaysOfYear (Day) as (
select DATEADD(year,DATEDIFF(year,0,CURRENT_TIMESTAMP),0) --Jan 1st
union all
select DATEADD(day,1,Day) from AllDaysOfYear
where DATEPART(year,DATEADD(day,1,Day)) = DATEPART(year,CURRENT_TIMESTAMP)
)
select
ROW_NUMBER() OVER (ORDER BY Day) as WeekNo,
Day
from
AllDaysOfYear
where
DATEPART(weekday,Day) = DATEPART(weekday,'20120714')
option (maxrecursion 0)
First, generate a set of all of the days in the current year (AllDaysInYear). Then, select those whose weekday is a saturday. The value I've used ('20120714') isn't terribly important - it just has to be any saturday, from any year. I'm just using it to avoid needing to have particular DATEFIRST or language settings.
This query shows how to get the first day of this year and the first day of the next year in the first part. The first day of the next year is calculated once so as not to keep getting and comparing the year parts.
;WITH cte(TheDate,NextYear) AS
(
SELECT CAST(CONVERT(CHAR(4),GETDATE(),112)+'0101' AS DATETIME),
CAST(YEAR(GETDATE())*10000+10101 AS CHAR(8))
UNION ALL
SELECT DateAdd(d,1,TheDate),NextYear
FROM cte
WHERE DateAdd(d,1,TheDate)<NextYear
)
SELECT Week = DatePart(wk,TheDate),
TheDate
FROM cte
WHERE DateName(dw,TheDate) in ('Saturday')
ORDER BY TheDate
OPTION (MAXRECURSION 366)
with t as
(
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
union all
select 1 b
)
select * from
(
select
current_timestamp
-datepart(dy,current_timestamp)
+row_number() over (order by t.b) d
from t, t t1, t t2
) tmp
where datepart(yyyy,d)=datepart(yyyy,current_timestamp)
and
DATENAME(dw,d)='sunday'
DECLARE #Year AS INT
SELECT #Year = 2020
;WITH weekends AS (
SELECT DATEFROMPARTS(#Year, 1, 1) AS dt
UNION ALL
SELECT DATEADD(DAY, 1, dt)
FROM weekends
WHERE dt < DATEFROMPARTS(#Year, 12, 31)
)
SELECT dt, DATENAME(MONTH, dt), DATENAME(DW, dt)
FROM weekends
WHERE DATEPART(DW, dt) IN (1, 7)
OPTION(MaxRecursion 366)

Getting Number of weeks in a Month from a Datetime Column

I have a table called FcData and the data looks like:
Op_Date
2011-02-14 11:53:40.000
2011-02-17 16:02:19.000
2010-02-14 12:53:40.000
2010-02-17 14:02:19.000
I am looking to get the Number of weeks in That Month from Op_Date. So I am looking for output like:
Op_Date Number of Weeks
2011-02-14 11:53:40.000 5
2011-02-17 16:02:19.000 5
2010-02-14 12:53:40.000 5
2010-02-17 14:02:19.000 5
This page has some good functions to figure out the last day of any given month: http://www.sql-server-helper.com/functions/get-last-day-of-month.aspx
Just wrap the output of that function with a DATEPART(wk, last_day_of_month) call. Combining it with an equivalent call for the 1st-day-of-week will let you get the number of weeks in that month.
Use this to get the number of week for ONE specific date. Replace GetDate() by your date:
declare #dt date = cast(GetDate() as date);
declare #dtstart date = DATEADD(day, -DATEPART(day, #dt) + 1, #dt);
declare #dtend date = dateadd(DAY, -1, DATEADD(MONTH, 1, #dtstart));
WITH dates AS (
SELECT #dtstart ADate
UNION ALL
SELECT DATEADD(day, 1, t.ADate)
FROM dates t
WHERE DATEADD(day, 1, t.ADate) <= #dtend
)
SELECT top 1 DatePart(WEEKDAY, ADate) weekday, COUNT(*) weeks
FROM dates d
group by DatePart(WEEKDAY, ADate)
order by 2 desc
Explained: the CTE creates a result set with all dates for the month of the given date. Then we query the result set, grouping by week day and count the number of occurrences. The max number will give us how many weeks the month overlaps (premise: if the month has 5 Mondays, it will cover five weeks of the year).
Update
Now, if you have multiple dates, you should tweak accordingly, joining your query with the dates CTE.
Here is my take on it, might have missed something.
In Linq:
from u in TblUsers
let date = u.CreateDate.Value
let firstDay = new DateTime(date.Year, date.Month, 1)
let lastDay = firstDay.AddMonths(1)
where u.CreateDate.HasValue
select Math.Ceiling((lastDay - firstDay).TotalDays / 7)
And generated SQL:
-- Region Parameters
DECLARE #p0 Int = 1
DECLARE #p1 Int = 1
DECLARE #p2 Float = 7
-- EndRegion
SELECT CEILING(((CONVERT(Float,CONVERT(BigInt,(((CONVERT(BigInt,DATEDIFF(DAY, [t3].[value], [t3].[value2]))) * 86400000) + DATEDIFF(MILLISECOND, DATEADD(DAY, DATEDIFF(DAY, [t3].[value], [t3].[value2]), [t3].[value]), [t3].[value2])) * 10000))) / 864000000000) / #p2) AS [value]
FROM (
SELECT [t2].[createDate], [t2].[value], DATEADD(MONTH, #p1, [t2].[value]) AS [value2]
FROM (
SELECT [t1].[createDate], CONVERT(DATETIME, CONVERT(NCHAR(2), DATEPART(Month, [t1].[value])) + ('/' + (CONVERT(NCHAR(2), #p0) + ('/' + CONVERT(NCHAR(4), DATEPART(Year, [t1].[value]))))), 101) AS [value]
FROM (
SELECT [t0].[createDate], [t0].[createDate] AS [value]
FROM [tblUser] AS [t0]
) AS [t1]
) AS [t2]
) AS [t3]
WHERE [t3].[createDate] IS NOT NULL
According to this MSDN article: http://msdn.microsoft.com/en-us/library/ms174420.aspx you can only get the current week in the year, not what that month returns.
There may be various approaches to implementing the idea suggested by #Marc B. Here's one, where no UDFs are used but the first and the last days of month are calculated directly:
WITH SampleData AS (
SELECT CAST('20110214' AS datetime) AS Op_Date
UNION ALL SELECT '20110217'
UNION ALL SELECT '20100214'
UNION ALL SELECT '20100217'
UNION ALL SELECT '20090214'
UNION ALL SELECT '20090217'
),
MonthStarts AS (
SELECT
Op_Date,
MonthStart = DATEADD(DAY, 1 - DAY(Op_Date), Op_Date)
/* alternatively: DATEADD(MONTH, DATEDIFF(MONTH, 0, Op_Date), 0) */
FROM FcData
),
Months AS (
SELECT
Op_Date,
MonthStart,
MonthEnd = DATEADD(DAY, -1, DATEADD(MONTH, 1, MonthStart))
FROM FcData
)
Weeks AS (
SELECT
Op_Date,
StartWeek = DATEPART(WEEK, MonthStart),
EndWeek = DATEPART(WEEK, MonthEnd)
FROM MonthStarts
)
SELECT
Op_Date,
NumberOfWeeks = EndWeek - StartWeek + 1
FROM Weeks
All calculations could be done in one SELECT, but I chose to split them into steps and place every step in a separate CTE so it could be seen better how the end result was obtained.
You can get number of weeks per month using the following method.
Datepart(WEEK,
DATEADD(DAY,
-1,
DATEADD(MONTH,
1,
DATEADD(DAY,
1 - DAY(GETDATE()),
GETDATE())))
-
DATEADD(DAY,
1 - DAY(GETDATE()),
GETDATE())
+1
)
Here how you can get accurate amount of weeks:
DECLARE #date DATETIME
SET #date = GETDATE()
SELECT ROUND(cast(datediff(day, dateadd(day, 1-day(#date), #date), dateadd(month, 1, dateadd(day, 1-day(#date), #date))) AS FLOAT) / 7, 2)
With this code for Sep 2014 you'll get 4.29 which is actually true since there're 4 full weeks and 2 more days.