Filling table with weeks of the year skips first week of 2021 - sql-server

So, I am using the following snippet on a procedure, which fills me a temporary table with the first day of each year's week, the week number and month name.
However, when I reach week 53, of 2020, it jumps to week 2 of 2021. This happens because the first of january is in the so called week 53 (which is correct), but it should also be creating another row with the first week of january 2021 (even with the sunday as being in 2020, as it should).
Snippet:
SET DATEFIRST 7
DECLARE #tblSundays TABLE (
[year] INT
,[month] INT
,[week] INT
,[date] DATETIME
)
DECLARE #DateFrom DATETIME = '2020-12-12'
,#DateTo DATETIME = '2021-06-06';
--select #DateFrom,#DateTo;
WITH CTE (dt)
AS (
SELECT #DateFrom
UNION ALL
SELECT DATEADD(d, 1, dt)
FROM CTE
WHERE dt < #DateTo
)
INSERT INTO #tblSundays
SELECT datepart(year, dt)
,datepart(month, dt)
,datepart(week, dt)
,dt
FROM CTE
WHERE datepart("dw", dt) = 1
OPTION (MAXRECURSION 1000)
;
select * from #tblSundays
Is there any way that I can do this within this snippet, or should I create a manual verification?
Thanks!

You should do a couple of little tricks to set week 1 to the first Sunday of the year (tested for the next 30 years):
set datefirst 7
declare #tblSundays table ([year] int, [month] int, [week] int, [date] date)
declare #DateFrom datetime = '20201212', #DateTo datetime = '20510112';
-- Get the next closest Sunday
select #DateFrom = dateAdd(wk, dateDiff(wk, 0, #datefrom - 1 ), 0) + 6
with CTE (dt) as (
select #DateFrom
union all
select dateAdd(dd, 7, dt) from CTE where dt < #DateTo
)
insert into #tblSundays
select datePart(yy, dt),
datePart(mm, dt),
datePart(wk, dt - (1 + (datePart(dy, dt) + 5) % 7) % 7),
dt
from CTE
-- where datepart(dw, dt) = 1
option (maxrecursion 1600);
select *, datePart(wk, [date]) as [standard_wk]
from #tblSundays

I don't see how you could have those two condition on that same snippet, but you could add a second query there to complete what you are trying to do.
Something like this, maybe:
SET datefirst 7
DECLARE #tblSundays TABLE
(
[year] INT,
[month] INT,
[week] INT,
[date] DATETIME
)
DECLARE #DateFrom DATETIME = '2020-12-12',
#DateTo DATETIME = '2021-06-06';
--select #DateFrom,#DateTo;
WITH cte (dt)
AS (SELECT #DateFrom
UNION ALL
SELECT Dateadd(d, 1, dt)
FROM cte
WHERE dt < #DateTo)
INSERT INTO #tblSundays
SELECT Datepart(year, dt),
Datepart(month, dt),
Datepart(week, dt),
dt
FROM cte
WHERE Datepart("dw", dt) = 1
OPTION (maxrecursion 1000);
--second new query
WITH cte (dt)
AS (SELECT #DateFrom
UNION ALL
SELECT Dateadd(d, 1, dt)
FROM cte
WHERE dt < #DateTo)
INSERT INTO #tblSundays
SELECT Datepart(year, dt),
Datepart(month, dt),
Datepart(week, dt),
dt
FROM cte
WHERE Datepart("dw", dt) <> 1
AND Datepart(day, dt) = 1
AND Datepart(week, dt) NOT IN (SELECT week
FROM #tblSundays)
OPTION (maxrecursion 1000);
SELECT *
FROM #tblSundays
ORDER BY year,
month,
week
Would that be OK?

Related

How to add dates within a range

I need to insert a range of dates into a table, along with corresponding column data. My table's earliest date stops at 2017-3-16, but I need to add dates going back to 2016-1-1. See screenshot below for reference:
I'm sure I can figure out how to attribute day of the week, type of day, etc. using functions such as datepart, datename, etc. What I'm not certain of is how to insert a range of dates, between 2016-1-1 and 2017-3-15.
I keep this snippet handy as it is often needed. Simply set the high and low date variables and join your date on the CalendarDate field. If date gaps are ok then INNER JOIN, otherwise LEFT JOIN.
DECLARE #StartDate DATETIME = '01/01/2015'
DECLARE #EndDate DATETIME = '12/01/2016'
;WITH OrderedDays as
(
SELECT CalendarDate = #StartDate
UNION ALL
SELECT CalendarDate = DATEADD(DAY, 1, CalendarDate)
FROM OrderedDays WHERE DATEADD (DAY, 1, CalendarDate) <= #EndDate
),
Calendar AS
(
SELECT
DayIndex = ROW_NUMBER() OVER(PARTITION BY 1 ORDER BY CalendarDate),
CalendarDate,
CalenderDayOfMonth = DATEPART(DAY, CalendarDate),
CalenderMonthOfYear = DATEPART(MONTH, CalendarDate),
CalendarYear = DATEPART(YEAR, CalendarDate),
CalenderWeekOfYear = DATEPART(WEEK, CalendarDate),
CalenderQuarterOfYear = DATEPART(QUARTER, CalendarDate),
CalenderDayOfYear = DATEPART(DAYOFYEAR, CalendarDate),
CalenderDayOfWeek = DATEPART(WEEKDAY, CalendarDate),
CalenderWeekday = DATENAME(WEEKDAY, CalendarDate)
FROM
OrderedDays
)
SELECT * FROM Calendar
OPTION (MAXRECURSION 0)

Generate counts of records for each day in a week

I use SQL Server 2014 for my project. I have the following code to produce the number of registrations of each day:
SELECT
DATEADD(DAY, DATEDIFF(DAY, 0, createTime), 0) AS createdOn,
COUNT(*) AS Count
FROM
Registration
GROUP BY
DATEADD(DAY, DATEDIFF(DAY, 0, createTime), 0)
ORDER BY
createdOn
Now I would like to get the numbers for each day in a week (so there will be max 7 rows in output). How can I do it?
Here is the solution I have based on George's comment. Thank you, George!
SELECT
DATEPART(weekday, createTime) AS createdOn,
COUNT(*) AS Count
FROM
Registration
GROUP BY
DATEPART(weekday, createTime)
ORDER BY
createdOn
One way to return all days within a range returned with your data joined on matching days is to use a "calendar" table and LEFT JOIN your data by date.
DECLARE #StartDate DATETIME = '01/01/2015'
DECLARE #EndDate DATETIME = '12/01/2016'
//By Day In Year
;WITH Calender as
(
SELECT CalendarDate = #StartDate
UNION ALL
SELECT CalendarDate = DATEADD(DAY, 1, CalendarDate)
FROM Calender WHERE DATEADD (DAY, 1, CalendarDate) <= #EndDate
)
SELECT
C.CalendarDate,
COUNT(*) AS Count
FROM
Calender C
LEFT JOIN Regsitration R ON R.createdOn = C.CalendarDate
GROUP BY
C.CalendarDate
OPTION (MAXRECURSION 0)
//By Week In Year
;WITH Calender as
(
SELECT CalendarDate = #StartDate, WeekNumber=DATEPART(WEEK, #StartDate)
UNION ALL
SELECT CalendarDate = DATEADD(WEEK, 1, CalendarDate), WeekNumber=DATEPART(WEEK, #StartDate)
FROM Calender WHERE DATEADD (WEEK, 1, CalendarDate) <= #EndDate
)
SELECT
C.WeekNumber,
COUNT(*) AS Count
FROM
Calender C
LEFT JOIN Regsitration R ON DATEPART(WEEK,R.createdOn) = C.WeekNumber
GROUP BY
C.WeekNumber
OPTION (MAXRECURSION 0)

How to get date from day name?

How can I get the date of specific day ? Like if I have Thursday or month number ?
If I give 12 for instance I want to get the date of 12th day of this month. Or if I give 'Sun' or 'Sat' is it possible to get the dates of these days ?
DATEFROMPARTS function can construct a date from day, month and year.
DATEPARTS does the opposite - gives you the day, month, year, hour, etc. of a date. Or you can use functions like YEAR, MONTH and DAY.
You can deconstruct the value returned by GETDATE function and construct whatever date you want. Here is for example how to get the date for 12th day of the current month:
select DATEFROMPARTS(YEAR(GETDATE()), MONTH(GETDATE()), 12)
Converting 'Sun' or 'Sat' to date is a bit more difficult. First, they aren't quite deterministic. If today is Friday, "Sunday this week" means "next Sunday" in some parts of the world and "last Sunday" in others. You should implement your own logic based on the value returned by DATEPART(dw, GETDATE()) (which will give you the day of the week).
To find the weekday of the current month
DECLARE #daynumber INT = 12
SELECT datename(weekday, dateadd(d, #daynumber - 1, getdate()))
To find the dates of the current month of a given weekday
DECLARE #dayname char(3) = 'sat'
;WITH CTE as
(
SELECt TOP
(datediff(D, eomonth(getdate(), -1),eomonth(getdate())))
dateadd(d,row_number()over(ORDER BY 1/0),
eomonth(getdate(),-1))date
FROM
(values(1),(2),(3),(4),(5),(6))x(x),
(values(1),(2),(3),(4),(5),(6))y(x)
)
SELECT day(date) monthday, date
FROM CTE
WHERE left(datename(weekday, date),3) = #dayname
select sysdatetime(); --2018-12-13 16:29:56.0560574
---If I give 12 for instance I want to get the date of 12th day of this month.
declare #numDate int = 12;
select dateadd(m, datediff(m,0,getdate()),#numDate - 1 ); --2018-12-12 00:00:00.000
--Or if I give 'Sun' or 'Sat' is it possible to get the dates of these days ?
declare #text nvarchar(20) = 'Sunday';
declare #dateStart date = dateadd(month, datediff(month, 0, sysdatetime()), 0),
#days int =( select (DAY(dateadd(dd,-1,DATEADD(m,1,cast(2018 as varchar(4)) + '-' + cast(12 as varchar(2)) +'-01')))));
declare #dateEnd date = DATEADD(day,#days-1,#dateStart);
;WITH CTE (Dates,EndDate) AS
(
SELECT #dateStart AS Dates,#dateEnd AS EndDate
UNION ALL
SELECT DATEADD(day,1,Dates),EndDate
FROM CTE
WHERE DATEADD(day,1,Dates) <= EndDate
)
SELECT CTE.Dates, DATENAME(DW, CTE.Dates)
FROM CTE
where DATENAME(DW, CTE.Dates) = #text;
Result:
Dates,Day
2018/12/2,Sunday
2018/12/9,Sunday
2018/12/16,Sunday
2018/12/23,Sunday
2018/12/30,Sunday
-- Here is how to get week day name to week day number
DECLARE #T TABLE (Dow INT, NameOfDay VARCHAR(15), ShortName CHAR(3));
WITH Days AS
(
SELECT TOP 7
ROW_NUMBER() OVER(PARTITION BY object_id ORDER BY object_id) AS RowNo
FROM
sys.all_columns
)
INSERT INTO #T
SELECT
RowNo,
DATENAME(WEEKDAY, RowNo - 1),
LEFT(DATENAME(WEEKDAY, RowNo - 1), 3)
FROM
Days
SELECT
*
FROM
#T;
-- Here is how to get start of period
SELECT
DATEADD(DAY, DATEDIFF(DAY, 0, GETDATE()), 0) AS StartOfDay,
DATEADD(WEEK, DATEDIFF(WEEK, 0, GETDATE()), 0) AS StartOfWeek,
DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0) AS StartOfMonth,
DATEADD(YEAR, DATEDIFF(YEAR, 0, GETDATE()), 0) AS StartOfYear;
-- An example
WITH
StartPeriods AS
(
SELECT DATEADD(WEEK, DATEDIFF(WEEK, 0, GETDATE()), 0) AS StartOfWeek
),
SelectedDay AS
(
SELECT
Dow - 1 AS Dow,
(SELECT StartOfWeek FROM StartPeriods) AS StartOfWeek
FROM
#T
WHERE
ShortName = 'Wed'
)
SELECT
DATEADD(DAY, Dow, StartOfWeek)
FROM
SelectedDay;

TSQL get last day of previous months upto a specified month

I need to get last day of all previous months including current month, upto a specified month. For example, I need last days of september, aug, july, june, may, april, march, feb, jan, dec 2015 like so:
temptable_mytable:
last_day_of_month
-----------------
2016-09-30
2016-08-31
2016-07-31
2016-06-30
2016-05-31
2016-04-30
2016-03-31
2016-02-30
2016-01-31
2015-12-31
I need to specify the month and year to go back to - in above case it's December 2015, but it could also be September 2015 and such. Is there a way that I can do a loop and do this instead of having to calculate separately for each month end?
Use a recursive CTE with the EOMONTH function.
DECLARE #startdate DATE = '2016-01-01'
;WITH CTE
AS
(
SELECT EOMONTH(GETDATE()) as 'Dates'
UNION ALL
SELECT EOMONTH(DATEADD(MONTH, -1, [Dates]))
FROM CTE WHERE Dates > DATEADD(MONTH, 1, #startdate)
)
SELECT * FROM CTE
with temp as (select -1 i union all
select i+1 i from temp where i < 8)
select DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE())+i*-1,0)) from temp
declare #LASTMONTH date = '2018-10-01';
WITH MTHS AS (
SELECT dateadd(month,month(getdate()),dateadd(year,year(getdate()) - 1900, 0)) aday
UNION ALL
SELECT DATEADD(month,1,aday) from MTHS WHERE aday <= #LASTMONTH
),
LASTDAYS AS (SELECT DATEADD(day,-1,aday) finaldayofmonth from MTHS)
select * from LASTDAYS
Here is a version that goes forward or backwards as appropriate
declare #LASTMONTH date = '2013-10-01';
WITH DIF AS (SELECT CASE WHEN
YEAR(#LASTMONTH) * 12 + MONTH(#LASTMONTH)
>= YEAR(GETDATE()) * 12 + MONTH(getdate()) THEN 1 ELSE -1 END x),
MTHS AS (
SELECT dateadd(month,month(getdate()),dateadd(year,year(getdate()) - 1900, 0)) aday
UNION ALL
SELECT DATEADD(month,(SELECT X from dif),aday) from MTHS
WHERE month(aday) != month(dateadd(month,1,#LASTMONTH)) or YEAR(aday) != YEAR(dateadd(month,1,#LASTMONTH))
),
LASTDAYS AS (SELECT DATEADD(day,-1,aday) finaldayofmonth from MTHS)
select * from LASTDAYS order by finaldayofmonth
Here's one approach, using a CTE to generate a list of incrementing numbers to allow us to then have something to select from and use in a DATEADD to go back for the appropriate number of months.
Typically, if you're doing this quite frequently, instead of generating numbers on the fly like this with the CROSS JOIN, I'd recommend just creating a "Numbers" table that just holds numbers from 1 to "some number high enough to meet your needs"
DECLARE #Date DATE = '20151201'
DECLARE #MonthsBackToGo INTEGER
SELECT #MonthsBackToGo = DATEDIFF(mm, #Date, GETDATE()) + 1;
WITH _Numbers AS
(
SELECT TOP (#MonthsBackToGo) ROW_NUMBER() OVER (ORDER BY o.object_id) AS Number
FROM sys.objects o
CROSS JOIN sys.objects o2
)
SELECT EOMONTH(DATEADD(mm, -(Number- 1), GETDATE())) AS last_day_of_month
FROM _Numbers
This should scale out no matter how far you go back or forward for your originating table or object.
SET NOCOUNT ON;
DECLARE #Dates TABLE ( dt DATE)
DECLARE #Start DATE = DATEADD(YEAR, DATEDIFF(YEAR, 0, GETDATE()), 0)
DECLARE #End DATE = DATEADD(YEAR, 1, #Start)
WHILE #Start <= #End
BEGIN
INSERT INTO #Dates (dt) VALUES (#Start)
SELECT #Start = DATEADD(DAY, 1, #Start)
END
; With x as
(
Select
dt
, ROW_NUMBER() OVER(PARTITION BY DATEPART(YEAR, Dt), DATEPART(MONTH, Dt) ORDER BY Dt Desc) AS rwn
From #Dates
)
Select *
From x
WHERE rwn = 1
ORDER BY Dt
This was cribbed together quick based on a couple different SO answers for the parts:
DECLARE #startdate datetime, #enddate datetime
set #startdate = '2015-12-01'
set #enddate = getdate()
;WITH T(date)
AS
(
SELECT #startdate
UNION ALL
SELECT DateAdd(day,1,T.date) FROM T WHERE T.date < #enddate
)
SELECT DISTINCT
DATEADD(
day,
-1,
CAST(CAST(YEAR(date) AS varchar) + '-' + CAST(MONTH(date)AS varchar) + '-01' AS DATETIME))
FROM T OPTION (MAXRECURSION 32767);

how to get the start and end dates of all weeks between two dates in SQL server?

I need to get all week start and end dates(weeks) between two dates and then run a query returning the number of records inserted in each of those weeks.
declare #sDate datetime,
#eDate datetime;
select #sDate = '2013-02-25',
#eDate = '2013-03-25';
--query to get all weeks between sDate and eDate
--query to return number of items inserted in each of the weeks returned
WEEK NoOfItems
-----------------------------------------
2013-02-25 5
2013-03-4 2
2013-03-11 7
You can use a recursive CTE to generate the list of dates:
;with cte as
(
select #sDate StartDate,
DATEADD(wk, DATEDIFF(wk, 0, #sDate), 6) EndDate
union all
select dateadd(ww, 1, StartDate),
dateadd(ww, 1, EndDate)
from cte
where dateadd(ww, 1, StartDate)<= #eDate
)
select *
from cte
See SQL Fiddle with Demo.
Then you can join this to your table, to return the additional details.
Here is my solution. Inspired by this answer
DECLARE #sDate DATE = DATEADD(MONTH, -6, GETDATE())
DECLARE #eDate DATE = GETDATE()
;WITH cte AS
(
SELECT DATEADD(WEEK, DATEDIFF(WEEK, 0, #sDate), 0) AS StartDate, DATEADD(WEEK, DATEDIFF(WEEK, 0, #sDate), 6) AS EndDate
UNION ALL
SELECT DATEADD(WEEK, 1, StartDate), DATEADD(WEEK, 1, EndDate)
FROM cte
WHERE DATEADD(WEEK, 1, StartDate) <= #eDate
)
SELECT * FROM cte

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