(Transact-SQL) DATEDIFF and leap years - sql

I have a problem with DATEDIFF function.
My date format is dd/mm/yyyy.
#START_DATE = 01/02/2004
#END_DATE = 29/01/2014
The query (DATEDIFF(DAY,#START_DATE,#END_DATE) / 365) return 10, but the number of correct years is 9. This happens because my query does not consider leap years.
What I can do to keep an accurate count?
Thanks.


I believe the following logic does what you want:
datediff(year,
#START_DATE - datepart(dayofyear, #START_DATE) + 1,
#END_DATE - datepart(dayofyear, #START_DATE) + 1
) as d2
Note: This treats that dates as datetime, because arithmetic is easier to express. You can also write this as:
datediff(year,
dateadd(day, - datepart(dayofyear, #START_DATE) + 1, #START_DATE),
dateadd(day, - datepart(dayofyear, #START_DATE) + 1, #END_DATE)
) as d2
The following query is a demonstration:
select datediff(year,
startdate - datepart(dayofyear, startdate) + 1,
enddate - datepart(dayofyear, startdate) + 1
) as d2
from (select cast('2004-02-01' as datetime) as startdate,
cast('2014-01-31' as datetime) as enddate
union all
select cast('2004-02-01' as datetime) as startdate,
cast('2014-02-01' as datetime) as enddate
) t


Technically there would be 365.242 days in a year, when accounting for leap years so:
FLOOR(DATEDIFF(day, #STARTDATE, #ENDDATE) / 365.242)
Should be more correct.
Test:
SELECT FLOOR(DATEDIFF(day, '1980-01-16','2015-01-15') / 365.242),
FLOOR(DATEDIFF(day, '1980-01-16','2015-01-16') / 365.242)
ResultSet:
--------------------------------------- ---------------------------------------
34 35
Cheers!


You can create a function to address that:
CREATE FUNCTION [dbo].[getYears]
(
#START_DATE datetime,
#END_DATE datetime
)
RETURNS int
AS
BEGIN
DECLARE #yrs int
SET #yrs =DATEDIFF(year,#START_DATE,#END_DATE)
IF (#END_DATE < DATEADD(year, #yrs, #START_DATE))
SET #yrs = #yrs -1
RETURN #yrs
END
Also check this


Count the number of leap days in end_date Deduct number of leap days in start_date. Deduct the answer from your DATEDIFF.
DECLARE
#START_DATE DATETIME = '2004-02-01',
#END_DATE DATETIME = '2014-01-29'
SELECT (
DATEDIFF(DAY,#START_DATE,#END_DATE)
- (
(CONVERT(INT,#END_DATE - 58) / 1461)
-
(CONVERT(INT,#START_DATE - 58) / 1461)
)
) / 365
-58 to ignore Jan and Feb 1900
/ 1461 being the number of days between leap years

Related

Datetime2 Overflow Issue [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;

how do I get the EXACT number of months between two dates?

DATEDIFF(MONTH, '1/1/2014', '12/31/2014') + 1
This will give me 12.
What if I do this:
DATEDIFF(MONTH, '1/1/2014', '12/30/2014') + 1
It should give me 11 point something. How do I go about getting the exact number of months between these two dates? This needs to work for any combination of dates (any month of the year for any year).

You could do the calculation yourself in the following way:
DECLARE #startdate date = '1/1/2014'
DECLARE #enddate date = '12/30/2014'
DECLARE #startday int = DATEPART(DAY, #startdate)
DECLARE #endday int = DATEPART(DAY, #enddate)
DECLARE #startdateBase date = DATEADD(DAY, 1 - #startday, #startdate)
DECLARE #enddateBase date = DATEADD(DAY, 1 - #endday, #enddate)
DECLARE #deciMonthDiff float = CAST(DATEDIFF(MONTH, #startdate, #enddate) AS float) -
(#startday - 1.0) / DATEDIFF(DAY, #startdateBase, DATEADD(MONTH, 1, #startdateBase)) +
(#endday - 1.0) / DATEDIFF(DAY, #enddateBase, DATEADD(MONTH, 1, #enddateBase))
SELECT #deciMonthDiff
This calculates the #deciMonthDiff to be 11.935483870967.
Of course you can "inline" this as much as you want in order to avoid all the middle declarations.
The idea is to calculate the total month diff, then subtract the relative part of the first & last month depending on the actual day.

DATEDIFF with the MONTH option only returns an integer value. Using days or years would give you a rough "guesstimate" but still not exactly right (different number of days in a month/year so you can't just divide the days difference by 30).
If you want exact you would need to write your own function to walk through the months from start until end and account for how many days are in each month and get a percentage/factor of that month covered.

DECLARE #startdate date = '1/1/2014'
DECLARE #enddate date = '12/30/2014'
select case when DATEPART(DAY, #startdate) <= DATEPART(DAY, #enddate)
then datediff(month, #startdate, #enddate)
else datediff(month, #startdate, #enddate) -1
end
from Chris The Jedi of T-SQL

How can i change this Dateadd to selective dates

In the end of my big procedure i have previously made it so i can write example: "2" to get sales information for February. But now i want to change this so i can write the specific dates, because we want to fetch some more data.
Example of what I want to write: 29.01.2014 - 28.02.2014
Code:
WHERE isa.sales_date >= Dateadd(mm, ( #year - 1900 ) * 12 + ( #month - 1 ),
0)
AND isa.sales_date < Dateadd(mm, ( #year - 1900 ) * 12 + #month, 0)

Try:
Have two parameters:
#startDate varchar(10) --2014-01-01
#endDate varchar(10) --2014-01-29
And modify code like:
WHERE isa.sales_date >= convert(datetime, #startDate, 120)
AND isa.sales_date < DATEADD(DAY, 1, (convert(datetime, #endDate, 120))
The DATEADD part is to include all sales until the end of the last day.

Date Value from a week number in SQL [duplicate]

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Closed 10 years ago.
Possible Duplicate:
Get dates from a week number in T-SQL
How do I get the date value if I have a week number in SQL Query.
Like if I pass 26, it should give me 06/24/2012. If I pass 27, I should get 07/01/2012
Any help will be appreciated :)
Sots

SELECT DATEADD(week, n, '11/25/2011');
with n being the week number

If this doesn't work, try using WEEK() instead of WEEKOFYEAR().
CURDATE() - INTERVAL WEEKDAY(CURDATE()) DAY + INTERVAL (WEEKNO - WEEKOFYEAR(CURDATE())) WEEK

In SQL Server
DECLARE #StartDate DATE, #WeekVal INT
SET #WeekVal = 26 -- Set the week number
SET #StartDate = DATEADD(YEAR, DATEDIFF(YEAR, 0, GETDATE()), 0) -- Start of current year
;WITH cte AS (
SELECT #StartDate AS DateVal, DATEPART(wk, #StartDate) AS WeekVal, 1 AS RowVal
UNION ALL
SELECT DATEADD(d, 1, DateVal), DATEPART(wk, DATEADD(d, 1, DateVal)), RowVal + 1
FROM cte WHERE RowVal < 365
)
SELECT MIN(DateVal) StartOfWeek
FROM cte
WHERE WeekVal = #WeekVal
OPTION (MAXRECURSION 365);

This gives you the start and end dates of the week. [For SQL Server]
Declare #week integer set #week = 26
Declare #Year Integer Set #Year = year(getdate())
declare #date datetime
-- ------------------------------------
Set #date = DateAdd(day, 0,
DateAdd(month, 0,
DateAdd(Year, #Year-1900, 0)))
set #date = Dateadd(week, #week-1, #date)
select #date startweek, DATEADD (D, -1 * DatePart (DW, #date) + 7, #date) endweek
This was the result from it:
startweek endweek
----------------------- -----------------------
2012-07-01 00:00:00.000 2012-07-07 00:00:00.000
(1 row(s) affected)

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.