Week start date and week end date calculated wrong - sql-server

I have a query for calculating first and last date in the week, according to given date. It is enough to set #dDate and the query will calculate first (monday) and last date (sunday) for that week.
Problem is, that is calculating wrong and I don't understand why.
Example:
#dDate = 2019-10-03 (year-month-day).
Result:
W_START W_END
2019-09-25 2019-10-01
But it should be:
2019-09-30 2019-10-06
Why is that?
Query:
set datefirst 1
declare #dDate date = cast('2019-10-16' as date)
select #dDAte
declare #year int = (select DATEPART(year, #dDAte))
select #year
declare #StartingDate date = cast(('' + cast(#year as nvarchar(4)) + '-01-01') as date)
select #StartingDate
declare #dateWeekEnd date = (select DATEADD(week, (datepart(week, cast(#dDate as date)) - 1), #StartingDate))
declare #dateWeekStart date = dateadd(day, -6, #dateWeekEnd)
select #dateWeekStart W_START, #dateWeekEnd W_END

Days of the week are so complicated. I find it easier to remember that 2001-01-01 fell on a Monday.
Then, the following date arithmetic does what you want:
select dateadd(day,
7 * (datediff(day, '2001-01-01', #dDate) / 7),
'2001-01-01' -- 2001-01-01 fell on a Monday
)
I admit this is something of a cop-out/hack. But SQL Server -- and other databases -- make such date arithmetic so cumbersome that simple tricks like this are handy to keep in mind.

Related

Find age between 15 to 25 years from a given column of date of birth? [duplicate]

I have a table listing people along with their date of birth (currently a nvarchar(25))
How can I convert that to a date, and then calculate their age in years?
My data looks as follows
ID Name DOB
1 John 1992-01-09 00:00:00
2 Sally 1959-05-20 00:00:00
I would like to see:
ID Name AGE DOB
1 John 17 1992-01-09 00:00:00
2 Sally 50 1959-05-20 00:00:00
There are issues with leap year/days and the following method, see the update below:
try this:
DECLARE #dob datetime
SET #dob='1992-01-09 00:00:00'
SELECT DATEDIFF(hour,#dob,GETDATE())/8766.0 AS AgeYearsDecimal
,CONVERT(int,ROUND(DATEDIFF(hour,#dob,GETDATE())/8766.0,0)) AS AgeYearsIntRound
,DATEDIFF(hour,#dob,GETDATE())/8766 AS AgeYearsIntTrunc
OUTPUT:
AgeYearsDecimal AgeYearsIntRound AgeYearsIntTrunc
--------------------------------------- ---------------- ----------------
17.767054 18 17
(1 row(s) affected)
UPDATE here are some more accurate methods:
BEST METHOD FOR YEARS IN INT
DECLARE #Now datetime, #Dob datetime
SELECT #Now='1990-05-05', #Dob='1980-05-05' --results in 10
--SELECT #Now='1990-05-04', #Dob='1980-05-05' --results in 9
--SELECT #Now='1989-05-06', #Dob='1980-05-05' --results in 9
--SELECT #Now='1990-05-06', #Dob='1980-05-05' --results in 10
--SELECT #Now='1990-12-06', #Dob='1980-05-05' --results in 10
--SELECT #Now='1991-05-04', #Dob='1980-05-05' --results in 10
SELECT
(CONVERT(int,CONVERT(char(8),#Now,112))-CONVERT(char(8),#Dob,112))/10000 AS AgeIntYears
you can change the above 10000 to 10000.0 and get decimals, but it will not be as accurate as the method below.
BEST METHOD FOR YEARS IN DECIMAL
DECLARE #Now datetime, #Dob datetime
SELECT #Now='1990-05-05', #Dob='1980-05-05' --results in 10.000000000000
--SELECT #Now='1990-05-04', #Dob='1980-05-05' --results in 9.997260273973
--SELECT #Now='1989-05-06', #Dob='1980-05-05' --results in 9.002739726027
--SELECT #Now='1990-05-06', #Dob='1980-05-05' --results in 10.002739726027
--SELECT #Now='1990-12-06', #Dob='1980-05-05' --results in 10.589041095890
--SELECT #Now='1991-05-04', #Dob='1980-05-05' --results in 10.997260273973
SELECT 1.0* DateDiff(yy,#Dob,#Now)
+CASE
WHEN #Now >= DATEFROMPARTS(DATEPART(yyyy,#Now),DATEPART(m,#Dob),DATEPART(d,#Dob)) THEN --birthday has happened for the #now year, so add some portion onto the year difference
( 1.0 --force automatic conversions from int to decimal
* DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,#Now),DATEPART(m,#Dob),DATEPART(d,#Dob)),#Now) --number of days difference between the #Now year birthday and the #Now day
/ DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,#Now),1,1),DATEFROMPARTS(DATEPART(yyyy,#Now)+1,1,1)) --number of days in the #Now year
)
ELSE --birthday has not been reached for the last year, so remove some portion of the year difference
-1 --remove this fractional difference onto the age
* ( -1.0 --force automatic conversions from int to decimal
* DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,#Now),DATEPART(m,#Dob),DATEPART(d,#Dob)),#Now) --number of days difference between the #Now year birthday and the #Now day
/ DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,#Now),1,1),DATEFROMPARTS(DATEPART(yyyy,#Now)+1,1,1)) --number of days in the #Now year
)
END AS AgeYearsDecimal
Gotta throw this one out there. If you convert the date using the 112 style (yyyymmdd) to a number you can use a calculation like this...
(yyyyMMdd - yyyyMMdd) / 10000 = difference in full years
declare #as_of datetime, #bday datetime;
select #as_of = '2009/10/15', #bday = '1980/4/20'
select
Convert(Char(8),#as_of,112),
Convert(Char(8),#bday,112),
0 + Convert(Char(8),#as_of,112) - Convert(Char(8),#bday,112),
(0 + Convert(Char(8),#as_of,112) - Convert(Char(8),#bday,112)) / 10000
output
20091015 19800420 290595 29
I have used this query in our production code for nearly 10 years:
SELECT FLOOR((CAST (GetDate() AS INTEGER) - CAST(Date_of_birth AS INTEGER)) / 365.25) AS Age
You need to consider the way the datediff command rounds.
SELECT CASE WHEN dateadd(year, datediff (year, DOB, getdate()), DOB) > getdate()
THEN datediff(year, DOB, getdate()) - 1
ELSE datediff(year, DOB, getdate())
END as Age
FROM <table>
Which I adapted from here.
Note that it will consider 28th February as the birthday of a leapling for non-leap years e.g. a person born on 29 Feb 2020 will be considered 1 year old on 28 Feb 2021 instead of 01 Mar 2021.
So many of the above solutions are wrong DateDiff(yy,#Dob, #PassedDate) will not consider the month and day of both dates. Also taking the dart parts and comparing only works if they're properly ordered.
THE FOLLOWING CODE WORKS AND IS VERY SIMPLE:
create function [dbo].[AgeAtDate](
#DOB datetime,
#PassedDate datetime
)
returns int
with SCHEMABINDING
as
begin
declare #iMonthDayDob int
declare #iMonthDayPassedDate int
select #iMonthDayDob = CAST(datepart (mm,#DOB) * 100 + datepart (dd,#DOB) AS int)
select #iMonthDayPassedDate = CAST(datepart (mm,#PassedDate) * 100 + datepart (dd,#PassedDate) AS int)
return DateDiff(yy,#DOB, #PassedDate)
- CASE WHEN #iMonthDayDob <= #iMonthDayPassedDate
THEN 0
ELSE 1
END
End
EDIT: THIS ANSWER IS INCORRECT. I leave it in here as a warning to anyone tempted to use dayofyear, with a further edit at the end.
If, like me, you do not want to divide by fractional days or risk rounding/leap year errors, I applaud #Bacon Bits comment in a post above https://stackoverflow.com/a/1572257/489865 where he says:
If we're talking about human ages, you should calculate it the way
humans calculate age. It has nothing to do with how fast the earth
moves and everything to do with the calendar. Every time the same
month and day elapses as the date of birth, you increment age by 1.
This means the following is the most accurate because it mirrors what
humans mean when they say "age".
He then offers:
DATEDIFF(yy, #date, GETDATE()) -
CASE WHEN (MONTH(#date) > MONTH(GETDATE())) OR (MONTH(#date) = MONTH(GETDATE()) AND DAY(#date) > DAY(GETDATE()))
THEN 1 ELSE 0 END
There are several suggestions here involving comparing the month & day (and some get it wrong, failing to allow for the OR as correctly here!). But nobody has offered dayofyear, which seems so simple and much shorter. I offer:
DATEDIFF(year, #date, GETDATE()) -
CASE WHEN DATEPART(dayofyear, #date) > DATEPART(dayofyear, GETDATE()) THEN 1 ELSE 0 END
[Note: Nowhere in SQL BOL/MSDN is what DATEPART(dayofyear, ...) returns actually documented! I understand it to be a number in the range 1--366; most importantly, it does not change by locale as per DATEPART(weekday, ...) & SET DATEFIRST.]
EDIT: Why dayofyear goes wrong: As user #AeroX has commented, if the birth/start date is after February in a non leap year, the age is incremented one day early when the current/end date is a leap year, e.g. '2015-05-26', '2016-05-25' gives an age of 1 when it should still be 0. Comparing the dayofyear in different years is clearly dangerous. So using MONTH() and DAY() is necessary after all.
I believe this is similar to other ones posted here.... but this solution worked for the leap year examples 02/29/1976 to 03/01/2011 and also worked for the case for the first year.. like 07/04/2011 to 07/03/2012 which the last one posted about leap year solution did not work for that first year use case.
SELECT FLOOR(DATEDIFF(DAY, #date1 , #date2) / 365.25)
Found here.
Since there isn't one simple answer that always gives the correct age, here's what I came up with.
SELECT DATEDIFF(YY, DateOfBirth, GETDATE()) -
CASE WHEN RIGHT(CONVERT(VARCHAR(6), GETDATE(), 12), 4) >=
RIGHT(CONVERT(VARCHAR(6), DateOfBirth, 12), 4)
THEN 0 ELSE 1 END AS AGE
This gets the year difference between the birth date and the current date. Then it subtracts a year if the birthdate hasn't passed yet.
Accurate all the time - regardless of leap years or how close to the birthdate.
Best of all - no function.
I've done a lot of thinking and searching about this and I have 3 solutions that
calculate age correctly
are short (mostly)
are (mostly) very understandable.
Here are testing values:
DECLARE #NOW DATETIME = '2013-07-04 23:59:59'
DECLARE #DOB DATETIME = '1986-07-05'
Solution 1: I found this approach in one js library. It's my favourite.
DATEDIFF(YY, #DOB, #NOW) -
CASE WHEN DATEADD(YY, DATEDIFF(YY, #DOB, #NOW), #DOB) > #NOW THEN 1 ELSE 0 END
It's actually adding difference in years to DOB and if it is bigger than current date then subtracts one year. Simple right? The only thing is that difference in years is duplicated here.
But if you don't need to use it inline you can write it like this:
DECLARE #AGE INT = DATEDIFF(YY, #DOB, #NOW)
IF DATEADD(YY, #AGE, #DOB) > #NOW
SET #AGE = #AGE - 1
Solution 2: This one I originally copied from #bacon-bits. It's the easiest to understand but a bit long.
DATEDIFF(YY, #DOB, #NOW) -
CASE WHEN MONTH(#DOB) > MONTH(#NOW)
OR MONTH(#DOB) = MONTH(#NOW) AND DAY(#DOB) > DAY(#NOW)
THEN 1 ELSE 0 END
It's basically calculating age as we humans do.
Solution 3: My friend refactored it into this:
DATEDIFF(YY, #DOB, #NOW) -
CEILING(0.5 * SIGN((MONTH(#DOB) - MONTH(#NOW)) * 50 + DAY(#DOB) - DAY(#NOW)))
This one is the shortest but it's most difficult to understand. 50 is just a weight so the day difference is only important when months are the same. SIGN function is for transforming whatever value it gets to -1, 0 or 1. CEILING(0.5 * is the same as Math.max(0, value) but there is no such thing in SQL.
What about:
DECLARE #DOB datetime
SET #DOB='19851125'
SELECT Datepart(yy,convert(date,GETDATE())-#DOB)-1900
Wouldn't that avoid all those rounding, truncating and ofsetting issues?
Just check whether the below answer is feasible.
DECLARE #BirthDate DATE = '09/06/1979'
SELECT
(
YEAR(GETDATE()) - YEAR(#BirthDate) -
CASE WHEN (MONTH(GETDATE()) * 100) + DATEPART(dd, GETDATE()) >
(MONTH(#BirthDate) * 100) + DATEPART(dd, #BirthDate)
THEN 1
ELSE 0
END
)
select floor((datediff(day,0,#today) - datediff(day,0,#birthdate)) / 365.2425) as age
There are a lot of 365.25 answers here. Remember how leap years are defined:
Every four years
except every 100 years
except every 400 years
There are many answers to this question, but I think this one is close to the truth.
The datediff(year,…,…) function, as we all know, only counts the boundaries crossed by the date part, in this case the year. As a result it ignores the rest of the year.
This will only give the age in completed years if the year were to start on the birthday. It probably doesn’t, but we can fake it by adjusting the asking date back by the same amount.
In pseudopseudo code, it’s something like this:
adjusted_today = today - month(dob) + 1 - day(dob) + 1
age = year(adjusted_today - dob)
The + 1 is to allow for the fact that the month and day numbers start from 1 and not 0.
The reason we subtract the month and the day separately rather than the day of the year is because February has the annoying tendency to change its length.
The calculation in SQL is:
datediff(year,dob,dateadd(month,-month(dob)+1,dateadd(day,-day(dob)+1,today)))
where dob and today are presumed to be the date of birth and the asking date.
You can test this as follows:
WITH dates AS (
SELECT
cast('2022-03-01' as date) AS today,
cast('1943-02-25' as date) AS dob
)
select
datediff(year,dob,dateadd(month,-month(dob)+1,dateadd(day,-day(dob)+1,today))) AS age
from dates;
which gives you George Harrison’s age in completed years.
This is much cleaner than fiddling about with quarter days which will generally give you misleading values on the edges.
If you have the luxury of creating a scalar function, you can use something like this:
DROP FUNCTION IF EXISTS age;
GO
CREATE FUNCTION age(#dob date, #today date) RETURNS INT AS
BEGIN
SET #today = dateadd(month,-month(#dob)+1,#today);
SET #today = dateadd(day,-day(#dob)+1,#today);
RETURN datediff(year,#dob,#today);
END;
GO
Remember, you need to call dbo.age() because, well, Microsoft.
DECLARE #DOB datetime
set #DOB ='11/25/1985'
select floor(
( cast(convert(varchar(8),getdate(),112) as int)-
cast(convert(varchar(8),#DOB,112) as int) ) / 10000
)
source: http://beginsql.wordpress.com/2012/04/26/how-to-calculate-age-in-sql-server/
Try This
DECLARE #date datetime, #tmpdate datetime, #years int, #months int, #days int
SELECT #date = '08/16/84'
SELECT #tmpdate = #date
SELECT #years = DATEDIFF(yy, #tmpdate, GETDATE()) - CASE WHEN (MONTH(#date) > MONTH(GETDATE())) OR (MONTH(#date) = MONTH(GETDATE()) AND DAY(#date) > DAY(GETDATE())) THEN 1 ELSE 0 END
SELECT #tmpdate = DATEADD(yy, #years, #tmpdate)
SELECT #months = DATEDIFF(m, #tmpdate, GETDATE()) - CASE WHEN DAY(#date) > DAY(GETDATE()) THEN 1 ELSE 0 END
SELECT #tmpdate = DATEADD(m, #months, #tmpdate)
SELECT #days = DATEDIFF(d, #tmpdate, GETDATE())
SELECT Convert(Varchar(Max),#years)+' Years '+ Convert(Varchar(max),#months) + ' Months '+Convert(Varchar(Max), #days)+'days'
After trying MANY methods, this works 100% of the time using the modern MS SQL FORMAT function instead of convert to style 112. Either would work but this is the least code.
Can anyone find a date combination which does not work? I don't think there is one :)
--Set parameters, or choose from table.column instead:
DECLARE #DOB DATE = '2000/02/29' -- If #DOB is a leap day...
,#ToDate DATE = '2018/03/01' --...there birthday in this calculation will be
--0+ part tells SQL to calc the char(8) as numbers:
SELECT [Age] = (0+ FORMAT(#ToDate,'yyyyMMdd') - FORMAT(#DOB,'yyyyMMdd') ) /10000
CASE WHEN datepart(MM, getdate()) < datepart(MM, BIRTHDATE) THEN ((datepart(YYYY, getdate()) - datepart(YYYY, BIRTH_DATE)) -1 )
ELSE
CASE WHEN datepart(MM, getdate()) = datepart(MM, BIRTHDATE)
THEN
CASE WHEN datepart(DD, getdate()) < datepart(DD, BIRTHDATE) THEN ((datepart(YYYY, getdate()) - datepart(YYYY, BIRTHDATE)) -1 )
ELSE (datepart(YYYY, getdate()) - datepart(YYYY, BIRTHDATE))
END
ELSE (datepart(YYYY, getdate()) - datepart(YYYY, BIRTHDATE)) END
END
SELECT ID,
Name,
DATEDIFF(yy,CONVERT(DATETIME, DOB),GETDATE()) AS AGE,
DOB
FROM MyTable
How about this:
SET #Age = CAST(DATEDIFF(Year, #DOB, #Stamp) as int)
IF (CAST(DATEDIFF(DAY, DATEADD(Year, #Age, #DOB), #Stamp) as int) < 0)
SET #Age = #Age - 1
Try this solution:
declare #BirthDate datetime
declare #ToDate datetime
set #BirthDate = '1/3/1990'
set #ToDate = '1/2/2008'
select #BirthDate [Date of Birth], #ToDate [ToDate],(case when (DatePart(mm,#ToDate) < Datepart(mm,#BirthDate))
OR (DatePart(m,#ToDate) = Datepart(m,#BirthDate) AND DatePart(dd,#ToDate) < Datepart(dd,#BirthDate))
then (Datepart(yy, #ToDate) - Datepart(yy, #BirthDate) - 1)
else (Datepart(yy, #ToDate) - Datepart(yy, #BirthDate))end) Age
This will correctly handle the issues with the birthday and rounding:
DECLARE #dob datetime
SET #dob='1992-01-09 00:00:00'
SELECT DATEDIFF(YEAR, '0:0', getdate()-#dob)
Ed Harper's solution is the simplest I have found which never returns the wrong answer when the month and day of the two dates are 1 or less days apart. I made a slight modification to handle negative ages.
DECLARE #D1 AS DATETIME, #D2 AS DATETIME
SET #D2 = '2012-03-01 10:00:02'
SET #D1 = '2013-03-01 10:00:01'
SELECT
DATEDIFF(YEAR, #D1,#D2)
+
CASE
WHEN #D1<#D2 AND DATEADD(YEAR, DATEDIFF(YEAR,#D1, #D2), #D1) > #D2
THEN - 1
WHEN #D1>#D2 AND DATEADD(YEAR, DATEDIFF(YEAR,#D1, #D2), #D1) < #D2
THEN 1
ELSE 0
END AS AGE
The answer marked as correct is nearer to accuracy but, it fails in following scenario - where Year of birth is Leap year and day are after February month
declare #ReportStartDate datetime = CONVERT(datetime, '1/1/2014'),
#DateofBirth datetime = CONVERT(datetime, '2/29/1948')
FLOOR(DATEDIFF(HOUR,#DateofBirth,#ReportStartDate )/8766)
OR
FLOOR(DATEDIFF(HOUR,#DateofBirth,#ReportStartDate )/8765.82) -- Divisor is more accurate than 8766
-- Following solution is giving me more accurate results.
FLOOR(DATEDIFF(YEAR,#DateofBirth,#ReportStartDate) - (CASE WHEN DATEADD(YY,DATEDIFF(YEAR,#DateofBirth,#ReportStartDate),#DateofBirth) > #ReportStartDate THEN 1 ELSE 0 END ))
It worked in almost all scenarios, considering leap year, date as 29 feb, etc.
Please correct me if this formula have any loophole.
Declare #dob datetime
Declare #today datetime
Set #dob = '05/20/2000'
set #today = getdate()
select CASE
WHEN dateadd(year, datediff (year, #dob, #today), #dob) > #today
THEN datediff (year, #dob, #today) - 1
ELSE datediff (year, #dob, #today)
END as Age
Here is how i calculate age given a birth date and current date.
select case
when cast(getdate() as date) = cast(dateadd(year, (datediff(year, '1996-09-09', getdate())), '1996-09-09') as date)
then dateDiff(yyyy,'1996-09-09',dateadd(year, 0, getdate()))
else dateDiff(yyyy,'1996-09-09',dateadd(year, -1, getdate()))
end as MemberAge
go
CREATE function dbo.AgeAtDate(
#DOB datetime,
#CompareDate datetime
)
returns INT
as
begin
return CASE WHEN #DOB is null
THEN
null
ELSE
DateDiff(yy,#DOB, #CompareDate)
- CASE WHEN datepart(mm,#CompareDate) > datepart(mm,#DOB) OR (datepart(mm,#CompareDate) = datepart(mm,#DOB) AND datepart(dd,#CompareDate) >= datepart(dd,#DOB))
THEN 0
ELSE 1
END
END
End
GO
DECLARE #FromDate DATETIME = '1992-01-2623:59:59.000',
#ToDate DATETIME = '2016-08-10 00:00:00.000',
#Years INT, #Months INT, #Days INT, #tmpFromDate DATETIME
SET #Years = DATEDIFF(YEAR, #FromDate, #ToDate)
- (CASE WHEN DATEADD(YEAR, DATEDIFF(YEAR, #FromDate, #ToDate),
#FromDate) > #ToDate THEN 1 ELSE 0 END)
SET #tmpFromDate = DATEADD(YEAR, #Years , #FromDate)
SET #Months = DATEDIFF(MONTH, #tmpFromDate, #ToDate)
- (CASE WHEN DATEADD(MONTH,DATEDIFF(MONTH, #tmpFromDate, #ToDate),
#tmpFromDate) > #ToDate THEN 1 ELSE 0 END)
SET #tmpFromDate = DATEADD(MONTH, #Months , #tmpFromDate)
SET #Days = DATEDIFF(DAY, #tmpFromDate, #ToDate)
- (CASE WHEN DATEADD(DAY, DATEDIFF(DAY, #tmpFromDate, #ToDate),
#tmpFromDate) > #ToDate THEN 1 ELSE 0 END)
SELECT #FromDate FromDate, #ToDate ToDate,
#Years Years, #Months Months, #Days Days
What about a solution with only date functions, not math, not worries about leap year
CREATE FUNCTION dbo.getAge(#dt datetime)
RETURNS int
AS
BEGIN
RETURN
DATEDIFF(yy, #dt, getdate())
- CASE
WHEN
MONTH(#dt) > MONTH(GETDATE()) OR
(MONTH(#dt) = MONTH(GETDATE()) AND DAY(#dt) > DAY(GETDATE()))
THEN 1
ELSE 0
END
END
declare #birthday as datetime
set #birthday = '2000-01-01'
declare #today as datetime
set #today = GetDate()
select
case when ( substring(convert(varchar, #today, 112), 5,4) >= substring(convert(varchar, #birthday, 112), 5,4) ) then
(datepart(year,#today) - datepart(year,#birthday))
else
(datepart(year,#today) - datepart(year,#birthday)) - 1
end
The following script checks the difference in years between now and the given date of birth; the second part checks whether the birthday is already past in the current year; if not, it subtracts it:
SELECT year(NOW()) - year(date_of_birth) - (CONCAT(year(NOW()), '-', month(date_of_birth), '-', day(date_of_birth)) > NOW()) AS Age
FROM tableName;

Get date from week and year - date must be first day of the year - sql

I have a query that converts week and year to date.
But it returns the exact date.
But what i want is, I need the date to be such a day, that is same as the first day of the year.
dateadd (week, PromisedWeek-1, dateadd (year, PromisedYear-1900, 0)) - 4 -
datepart(dw, dateadd (week, PromisedWeek-1, dateadd (year, PromisedYear-1900, 0)) - 4) + 1
Hypothetical example.
My current query does is:
If week is 4 and year 2017, returns 26-Sun-2017
My need:
If week is 4 and year 2017 and January 1st was a Wednesday, its should return 29-Wednesday-2017.
Hoping that you guys get what I am trying to explain.
I need the query to return such a date which has the same day as that of the current year's 1st day.
Unless I'm reading this wrong, you're making things too hard on yourself. Just do this:
declare #year int; set #year = 2017;
declare #week int; set #week = 4;
select dateadd(week, #week, dateadd(year, #year - 1900, 0))
Result:
2017-01-29 00:00:00.000
The inner dateadd() gives you the first day of the requested year (Jan 1), regardless of weekday. If it's a Wednesday, you'll get a Wednesday. This year, it's Sunday. The outer dateadd() then adds full weeks to that, so you end up with the same day of the week, just like you asked.
If that's not what you want, please explain how it needs to be more complicated.
This is pretty easy if you have a calendar table. A calendar table is a table that contains every date and its parts for a given range. I created mine for 2010 to 2020 with this code:
declare #start_dt as date = '1/1/2010';
declare #end_dt as date = '1/1/2020';
declare #dates as table (
date_id date primary key,
date_year smallint,
date_month tinyint,
date_day tinyint,
weekday_id tinyint,
weekday_nm varchar(10),
month_nm varchar(10),
day_of_year smallint,
quarter_id tinyint,
first_day_of_month date,
last_day_of_month date,
start_dts datetime,
end_dts datetime
)
while #start_dt < #end_dt
begin
insert into #dates(
date_id, date_year, date_month, date_day,
weekday_id, weekday_nm, month_nm, day_of_year, quarter_id,
first_day_of_month, last_day_of_month,
start_dts, end_dts
)
values(
#start_dt, year(#start_dt), month(#start_dt), day(#start_dt),
datepart(weekday, #start_dt), datename(weekday, #start_dt), datename(month, #start_dt), datepart(dayofyear, #start_dt), datepart(quarter, #start_dt),
dateadd(day,-(day(#start_dt)-1),#start_dt), dateadd(day,-(day(dateadd(month,1,#start_dt))),dateadd(month,1,#start_dt)),
cast(#start_dt as datetime), dateadd(second,-1,cast(dateadd(day, 1, #start_dt) as datetime))
)
set #start_dt = dateadd(day, 1, #start_dt)
end
select *
into Calendar
from #dates
Once you have a calendar table you can query for the correct weekday in the PromisedWeek
declare #PromisedYear as numeric
declare #PromisedWeek as int
set #PromisedYear = 2017
set #PromisedWeek = 4
select concat(date_day, '-', weekday_nm, '-', #PromisedYear)
from Calendar as c
where
weekday_id =
(select weekday_ID
from Calendar
where date_year = #PromisedYear
and day_of_year = 1
)
and datepart(week, date_id) = #PromisedWeek
and date_year = #PromisedYear
Turns out that January 1st 2017 was a Sunday and the 4th Sunday in 2017 was January 22nd, so the return is 22-Sunday-2017

Check if date falls within Month and Year range

If a user selects a range such as:
Start: November 2016
End: September 2017,
I want to include all results that fall within the range of 2016-11-01 to 2017-09-30.
I tried concatenating together the year, month, and day, however the issue comes that not all months have the same last day. While I know all months start on day 01, a month's end day can be 28, 29, 30, or 31.
Is there a way to do this without constructing the date? SqlServer 2008 doesn't have the EOMONTH function, and I feel like anything more complex than that is not the right solution. I would like to avoid this:
WHERE
DateCol >= '2016' + '-' + '11' + '-01' AND
DateCol <= '2017' + '-' + '09' + '-30'
It really seems to me that the easiest and best answer is to go from the first of the beginning month to the first of the month after the ending month, and make the second comparison not inclusive.
In other words, instead of this:
WHERE
DateCol >= '2016' + '-' + '11' + '-01' AND
DateCol <= '2017' + '-' + '09' + '-30'
simply this:
WHERE
DateCol >= '2016' + '-' + '11' + '-01' AND
DateCol < '2017' + '-' + '10' + '-01'
There is a faster way to do so :
DECLARE #minDate DATE
DECLARE #maxDate DATE
SET #minDate = XXXXX
SET #maxDate = YYYYY
-- Get the first day of the month minDate.
SET #minDate = CONVERT(datetime,CONVERT(varchar(6),#minDate,112)+'01',112)
-- Get the last day of the month minDate.
SET #maxDate = CONVERT(datetime,CONVERT(varchar(6),#maxDate,112)+'01',112)
SET #maxDate = DATEADD(day, -1, DATEADD(month, 1, #maxDate))
SELECT * FROM myTABLE WHERE DateCol >= #minDate AND DateCol <= #maxDate
Or :
SELECT * FROM myTABLE
WHERE DateCol >= CONVERT(datetime,CONVERT(varchar(6),XXXXX,112)+'01',112)
AND DateCol <= DATEADD(day, -1, DATEADD(month, 1, CONVERT(datetime,CONVERT(varchar(6),YYYYY,112)+'01',112)))
Use syntax like CONVERT(datetime,'20170930',112) or CONVERT(datetime,'09-30-2017',110) for XXXXX and YYYYY rather than '2017-09-30' that use SQL Server implicit convertion from char to datetime (rely on the server configuration : can be hazardous!!!)).
Using this syntax is faster because #minDate and #maxDate do not need any evaluation. So that indexes can be used directly...
Otherwise a scalar function that will simulate the eomonth() behaviour could be usefull...
You could use following select statement to get last date of any month (and any year) by passing a field or date to it:
DECLARE #dtDate DATE
SET #dtDate = '09/25/2016'
SELECT CAST(DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#dtDate)+1,0)) AS DATE) AS LastDay_AnyMonth
Please provide some example data and desired result and I will update my answer further.
Here you go:
DECLARE #YourTable TABLE (YourData DATE);
INSERT INTO #YourTable VALUES
('2016-11-01'),
('2016-09-05'),
('2017-03-03'),
('2017-11-11'),
('2017-12-14'),
('2017-09-30');
WITH CTE AS (
SELECT YourData
FROM #YourTable
WHERE YEAR(YourData) =2016 AND MONTH (YourData) >= 11
)
SELECT YourData
FROM #YourTable
WHERE YEAR(YourData) =2017 AND MONTH (YourData) <= 9
UNION ALL
SELECT YourData
FROM CTE;
There is no need to know the end of the month (28 or 30 or 31).
For 2008, you can simply convert the string
Example
Select Date1=convert(date,'November 2016')
,Date2=dateadd(DAY,-1,dateadd(MONTH,1,convert(date,'September 2017')))
Returns
Date1 Date2
2016-11-01 2017-09-30
So the WHERE would be somthing like this
...
Where DateCol between convert(date,'November 2016')
and dateadd(DAY,-1,dateadd(MONTH,1,convert(date,'September 2017')))
A useful construct for performing date-based operations is to make use of a Date Dimension table. You are creating a lookup table that is populated with a lot of information about dates over a large span of time. You can then query the table based on the information that you do have. The table is small enough so that it does not impose significant performance concerns.
In your particular case, you have the month and year. You would plug that into the date dimension table to get the first of the month from the beginning month and the last of the month from the ending month. You now have a time range to search over without any complex logic or calculations on the fly.
Aaron Bertrand explains it in depth here: https://www.mssqltips.com/sqlservertip/4054/creating-a-date-dimension-or-calendar-table-in-sql-server/

Select records created in a 24 hour time-frame

I'm trying to query our database to find all records that were created between 6am yesterday and 6am today. This will be run in a report at any point during the day so set times/dates are useless.
I have this so far:-
SELECT * FROM DaySummaryDetail DSD
WHERE DSD.FromDateTime BETWEEN DATEADD(DAY, -1, GetDate())
AND DATEADD(Day, 1, GetDate())
But obviously this only works for 24 hours ago from right now until right now. I can't figure out how to apply a time as well as date.
Every example I find online seems slightly different and uses set dates/times ie, >= 20/02/2015 06:00:00.
I normally use Oracle SQL which would simply work using this:-
ptt.mod_date_time >= TRUNC (SYSDATE - 1) - 2 / 24
AND ptt.mod_date_time <= TRUNC (SYSDATE - 1) + 22 / 24
This would return results from 10pm to 10pm but the format appears totally different in SQL Server.
You can get the datetime values you are after by doing the following:
SELECT DATEADD(HOUR,6,CONVERT(DATETIME, CONVERT(DATE ,GETDATE()))) Today6AM,
DATEADD(HOUR,-18,CONVERT(DATETIME, CONVERT(DATE ,GETDATE()))) Yesterday6AM
By doing this: CONVERT(DATE ,GETDATE()) you are stripping off the time portion of today's date. Converting it back to datetime gives you midnight for today.
The query adds 6 hours to midnight of the current day for 6am today and subtracts 18 hours from midnight of the current day to give you 6am on the previous day.
Output:
Today6AM Yesterday6AM
================================================
2015-02-20 06:00:00.000 2015-02-19 06:00:00.000
So adding that to your query:
SELECT *
FROM DaySummaryDetail DSD
WHERE DSD.FromDateTime
BETWEEN DATEADD(HOUR,-18,CONVERT(DATETIME, CONVERT(DATE ,GETDATE())))
AND DATEADD(HOUR,6,CONVERT(DATETIME, CONVERT(DATE ,GETDATE())))
DECLARE #StartTimestamp datetime
DECLARE #EndTimestamp datetime
DECLARE #HourPartOfSearchRange nvarchar(6)
SET #HourPartOfSearchRange = ' 06:30'
SET #StartTimestamp =
CAST((CONVERT(varchar(11), DATEADD(DAY,-1,#CurrentUTCDateTime), 106) + #HourPartOfSearchRange) AS datetime)
SET #EndTimestamp =
CAST((CONVERT(varchar(11), #CurrentUTCDateTime, 106) + #HourPartOfSearchRange) AS datetime)
SELECT * FROM dbo.Test Where Timestamp Between #StartTimestamp AND #EndTimestamp
today 6am is
dateadd(hour,6,cast(cast(getdate() as date) as datetime))
cast(getdate() as date) truncates the timepart, cast it back as datetime because dateadd won't add hours otherwise and add 6hours
One solution would be like so:
select *
from DaySummaryDetail DSD
where DSD.FromDateTime between cast(cast(cast(getdate()-1 as date) as varchar(30)) + ' 06:00:00.000' as datetime)
and cast(cast(cast(getdate() as date) as varchar(30)) + ' 06:00:00.000' as datetime)
This should help ...
SELECT DATEADD( hour, 6, CAST(CAST(GETDATE(), AS Date) AS DateTime) ) AS 'Today#6am'
SELECT DATEADD( hour, 6, CAST(CAST(GETDATE()-1, AS Date) AS DateTime) ) AS 'Yesterday#6am'
In SQL Server 2012 you can use SMALLDATETIMEFROMPARTS to construct a datetime value that is today at 6am like this:
SMALLDATETIMEFROMPARTS(YEAR(GETDATE()), MONTH(GETDATE()), DAY(GETDATE()), 6, 0)
Output: 2015-02-20 06:00:00
then you can use the above expression in place of GETDATE() in the WHERE clause:
DECLARE #TodayAt6AM DATETIME = SMALLDATETIMEFROMPARTS(YEAR(GETDATE()),
MONTH(GETDATE()),
DAY(GETDATE()),
6,
0)
SELECT *
FROM DaySummaryDetail DSD
WHERE DSD.FromDateTime BETWEEN DATEADD(DAY, -1, #TodayAt6AM) AND
DATEADD(Day, 1, #TodayAt6AM)

Get first day of week in SQL Server

I am trying to group records by week, storing the aggregated date as the first day of the week. However, the standard technique I use for rounding off dates does not appear to work correctly with weeks (though it does for days, months, years, quarters and any other timeframe I've applied it to).
Here is the SQL:
select "start_of_week" = dateadd(week, datediff(week, 0, getdate()), 0);
This returns 2011-08-22 00:00:00.000, which is a Monday, not a Sunday. Selecting ##datefirst returns 7, which is the code for Sunday, so the server is setup correctly in as far as I know.
I can bypass this easily enough by changing the above code to:
select "start_of_week" = dateadd(week, datediff(week, 0, getdate()), -1);
But the fact that I have to make such an exception makes me a little uneasy. Also, apologies if this is a duplicate question. I found some related questions but none that addressed this aspect specifically.
To answer why you're getting a Monday and not a Sunday:
You're adding a number of weeks to the date 0. What is date 0? 1900-01-01. What was the day on 1900-01-01? Monday. So in your code you're saying, how many weeks have passed since Monday, January 1, 1900? Let's call that [n]. Ok, now add [n] weeks to Monday, January 1, 1900. You should not be surprised that this ends up being a Monday. DATEADD has no idea that you want to add weeks but only until you get to a Sunday, it's just adding 7 days, then adding 7 more days, ... just like DATEDIFF only recognizes boundaries that have been crossed. For example, these both return 1, even though some folks complain that there should be some sensible logic built in to round up or down:
SELECT DATEDIFF(YEAR, '2010-01-01', '2011-12-31');
SELECT DATEDIFF(YEAR, '2010-12-31', '2011-01-01');
To answer how to get a Sunday:
If you want a Sunday, then pick a base date that's not a Monday but rather a Sunday. For example:
DECLARE #dt DATE = '1905-01-01';
SELECT [start_of_week] = DATEADD(WEEK, DATEDIFF(WEEK, #dt, CURRENT_TIMESTAMP), #dt);
This will not break if you change your DATEFIRST setting (or your code is running for a user with a different setting) - provided that you still want a Sunday regardless of the current setting. If you want those two answers to jive, then you should use a function that does depend on the DATEFIRST setting, e.g.
SELECT DATEADD(DAY, 1-DATEPART(WEEKDAY, CURRENT_TIMESTAMP), CURRENT_TIMESTAMP);
So if you change your DATEFIRST setting to Monday, Tuesday, what have you, the behavior will change. Depending on which behavior you want, you could use one of these functions:
CREATE FUNCTION dbo.StartOfWeek1 -- always a Sunday
(
#d DATE
)
RETURNS DATE
AS
BEGIN
RETURN (SELECT DATEADD(WEEK, DATEDIFF(WEEK, '19050101', #d), '19050101'));
END
GO
...or...
CREATE FUNCTION dbo.StartOfWeek2 -- always the DATEFIRST weekday
(
#d DATE
)
RETURNS DATE
AS
BEGIN
RETURN (SELECT DATEADD(DAY, 1-DATEPART(WEEKDAY, #d), #d));
END
GO
Now, you have plenty of alternatives, but which one performs best? I'd be surprised if there would be any major differences but I collected all the answers provided so far and ran them through two sets of tests - one cheap and one expensive. I measured client statistics because I don't see I/O or memory playing a part in the performance here (though those may come into play depending on how the function is used). In my tests the results are:
"Cheap" assignment query:
Function - client processing time / wait time on server replies / total exec time
Gandarez - 330/2029/2359 - 0:23.6
me datefirst - 329/2123/2452 - 0:24.5
me Sunday - 357/2158/2515 - 0:25.2
trailmax - 364/2160/2524 - 0:25.2
Curt - 424/2202/2626 - 0:26.3
"Expensive" assignment query:
Function - client processing time / wait time on server replies / total exec time
Curt - 1003/134158/135054 - 2:15
Gandarez - 957/142919/143876 - 2:24
me Sunday - 932/166817/165885 - 2:47
me datefirst - 939/171698/172637 - 2:53
trailmax - 958/173174/174132 - 2:54
I can relay the details of my tests if desired - stopping here as this is already getting quite long-winded. I was a bit surprised to see Curt's come out as the fastest at the high end, given the number of calculations and inline code. Maybe I'll run some more thorough tests and blog about it... if you guys don't have any objections to me publishing your functions elsewhere.
For these that need to get:
Monday = 1 and Sunday = 7:
SELECT 1 + ((5 + DATEPART(dw, GETDATE()) + ##DATEFIRST) % 7);
Sunday = 1 and Saturday = 7:
SELECT 1 + ((6 + DATEPART(dw, GETDATE()) + ##DATEFIRST) % 7);
Above there was a similar example, but thanks to double "%7" it would be much slower.
For those who need the answer at work and creating function is forbidden by your DBA, the following solution will work:
select *,
cast(DATEADD(day, -1*(DATEPART(WEEKDAY, YouDate)-1), YourDate) as DATE) as WeekStart
From.....
This gives the start of that week. Here I assume that Sundays are the start of weeks. If you think that Monday is the start, you should use:
select *,
cast(DATEADD(day, -1*(DATEPART(WEEKDAY, YouDate)-2), YourDate) as DATE) as WeekStart
From.....
This works wonderfully for me:
CREATE FUNCTION [dbo].[StartOfWeek]
(
#INPUTDATE DATETIME
)
RETURNS DATETIME
AS
BEGIN
-- THIS does not work in function.
-- SET DATEFIRST 1 -- set monday to be the first day of week.
DECLARE #DOW INT -- to store day of week
SET #INPUTDATE = CONVERT(VARCHAR(10), #INPUTDATE, 111)
SET #DOW = DATEPART(DW, #INPUTDATE)
-- Magic convertion of monday to 1, tuesday to 2, etc.
-- irrespect what SQL server thinks about start of the week.
-- But here we have sunday marked as 0, but we fix this later.
SET #DOW = (#DOW + ##DATEFIRST - 1) %7
IF #DOW = 0 SET #DOW = 7 -- fix for sunday
RETURN DATEADD(DD, 1 - #DOW,#INPUTDATE)
END
Maybe you need this:
SELECT DATEADD(DD, 1 - DATEPART(DW, GETDATE()), GETDATE())
Or
DECLARE #MYDATE DATETIME
SET #MYDATE = '2011-08-23'
SELECT DATEADD(DD, 1 - DATEPART(DW, #MYDATE), #MYDATE)
Function
CREATE FUNCTION [dbo].[GetFirstDayOfWeek]
( #pInputDate DATETIME )
RETURNS DATETIME
BEGIN
SET #pInputDate = CONVERT(VARCHAR(10), #pInputDate, 111)
RETURN DATEADD(DD, 1 - DATEPART(DW, #pInputDate),
#pInputDate)
END
GO
Googled this script:
create function dbo.F_START_OF_WEEK
(
#DATE datetime,
-- Sun = 1, Mon = 2, Tue = 3, Wed = 4
-- Thu = 5, Fri = 6, Sat = 7
-- Default to Sunday
#WEEK_START_DAY int = 1
)
/*
Find the fisrt date on or before #DATE that matches
day of week of #WEEK_START_DAY.
*/
returns datetime
as
begin
declare #START_OF_WEEK_DATE datetime
declare #FIRST_BOW datetime
-- Check for valid day of week
if #WEEK_START_DAY between 1 and 7
begin
-- Find first day on or after 1753/1/1 (-53690)
-- matching day of week of #WEEK_START_DAY
-- 1753/1/1 is earliest possible SQL Server date.
select #FIRST_BOW = convert(datetime,-53690+((#WEEK_START_DAY+5)%7))
-- Verify beginning of week not before 1753/1/1
if #DATE >= #FIRST_BOW
begin
select #START_OF_WEEK_DATE =
dateadd(dd,(datediff(dd,#FIRST_BOW,#DATE)/7)*7,#FIRST_BOW)
end
end
return #START_OF_WEEK_DATE
end
go
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=47307
CREATE FUNCTION dbo.fnFirstWorkingDayOfTheWeek
(
#currentDate date
)
RETURNS INT
AS
BEGIN
-- get DATEFIRST setting
DECLARE #ds int = ##DATEFIRST
-- get week day number under current DATEFIRST setting
DECLARE #dow int = DATEPART(dw,#currentDate)
DECLARE #wd int = 1+(((#dow+#ds) % 7)+5) % 7 -- this is always return Mon as 1,Tue as 2 ... Sun as 7
RETURN DATEADD(dd,1-#wd,#currentDate)
END
For the basic (the current week's Sunday)
select cast(dateadd(day,-(datepart(dw,getdate())-1),getdate()) as date)
If previous week:
select cast(dateadd(day,-(datepart(dw,getdate())-1),getdate()) -7 as date)
Internally, we built a function that does it but if you need quick and dirty, this will do it.
Since Julian date 0 is a Monday just add the number of weeks to Sunday
which is the day before -1 Eg. select dateadd(wk,datediff(wk,0,getdate()),-1)
I found some of the other answers long-winded or didn't actually work if you wanted Monday as the start of the week.
Sunday
SELECT DATEADD(week, DATEDIFF(week, -1, GETDATE()), -1) AS Sunday;
Monday
SELECT DATEADD(week, DATEDIFF(week, 0, GETDATE() - 1), 0) AS Monday;
Set DateFirst 1;
Select
Datepart(wk, TimeByDay) [Week]
,Dateadd(d,
CASE
WHEN Datepart(dw, TimeByDay) = 1 then 0
WHEN Datepart(dw, TimeByDay) = 2 then -1
WHEN Datepart(dw, TimeByDay) = 3 then -2
WHEN Datepart(dw, TimeByDay) = 4 then -3
WHEN Datepart(dw, TimeByDay) = 5 then -4
WHEN Datepart(dw, TimeByDay) = 6 then -5
WHEN Datepart(dw, TimeByDay) = 7 then -6
END
, TimeByDay) as StartOfWeek
from TimeByDay_Tbl
This is my logic. Set the first of the week to be Monday then calculate what is the day of the week a give day is, then using DateAdd and Case I calculate what the date would have been on the previous Monday of that week.
This is a useful function for me
/* MeRrais 211126
select [dbo].[SinceWeeks](0,NULL)
select [dbo].[SinceWeeks](5,'2021-08-31')
*/
alter Function [dbo].[SinceWeeks](#Weeks int, #From datetime=NULL)
Returns date
AS
Begin
if #From is null
set #From=getdate()
return cast(dateadd(day, -(#Weeks*7+datepart(dw,#From)-1), #From) as date)
END
I don't have any issues with any of the answers given here, however I do think mine is a lot simpler to implement, and understand. I have not run any performance tests on it, but it should be neglegable.
So I derived my answer from the fact that dates are stored in SQL server as integers, (I am talking about the date component only). If you don't believe me, try this SELECT CONVERT(INT, GETDATE()), and vice versa.
Now knowing this, you can do some cool math equations. You might be able to come up with a better one, but here is mine.
/*
TAKEN FROM http://msdn.microsoft.com/en-us/library/ms181598.aspx
First day of the week is
1 -- Monday
2 -- Tuesday
3 -- Wednesday
4 -- Thursday
5 -- Friday
6 -- Saturday
7 (default, U.S. English) -- Sunday
*/
--Offset is required to compensate for the fact that my ##DATEFIRST setting is 7, the default.
DECLARE #offSet int, #testDate datetime
SELECT #offSet = 1, #testDate = GETDATE()
SELECT CONVERT(DATETIME, CONVERT(INT, #testDate) - (DATEPART(WEEKDAY, #testDate) - #offSet))
I had a similar problem. Given a date, I wanted to get the date of the Monday of that week.
I used the following logic: Find the day number in the week in the range of 0-6, then subtract that from the originay date.
I used: DATEADD(day,-(DATEPART(weekday,)+5)%7,)
Since DATEPRRT(weekday,) returns 1 = Sundaye ... 7=Saturday,
DATEPART(weekday,)+5)%7 returns 0=Monday ... 6=Sunday.
Subtracting this number of days from the original date gives the previous Monday. The same technique could be used for any starting day of the week.
I found this simple and usefull. Works even if first day of week is Sunday or Monday.
DECLARE #BaseDate AS Date
SET #BaseDate = GETDATE()
DECLARE #FisrtDOW AS Date
SELECT #FirstDOW = DATEADD(d,DATEPART(WEEKDAY,#BaseDate) *-1 + 1, #BaseDate)
Maybe I'm over simplifying here, and that may be the case, but this seems to work for me. Haven't ran into any problems with it yet...
CAST('1/1/' + CAST(YEAR(GETDATE()) AS VARCHAR(30)) AS DATETIME) + (DATEPART(wk, YOUR_DATE) * 7 - 7) as 'FirstDayOfWeek'
CAST('1/1/' + CAST(YEAR(GETDATE()) AS VARCHAR(30)) AS DATETIME) + (DATEPART(wk, YOUR_DATE) * 7) as 'LastDayOfWeek'

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