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;
I have a need to create a gross requirements report that takes how much supply and demand of a item in inventory from a start date onwards and 'buckets' it into different weeks of the year so that material planners know when they will need a item and if they have enough stock in inventory at that time.
As an example, today’s date (report date) is 8/27/08. The first step is to find the date for the Monday of the week the report date falls in. In this case, Monday would be 8/25/08. This becomes the first day of the first bucket. All transactions that fall before that are assigned to week #0 and will be summarized as the beginning balance for the report. The remaining buckets are calculated from that point. For the eighth bucket, there is no ending date so any transactions after that 8th bucket start date are considered week #8.
WEEK# START DATE END DATE
0.......None..........8/24/08
1.......8/25/08.......8/31/08
2.......9/1/08.........9/7/08
3.......9/8/08.........9/14/08
4.......9/15/08.......9/21/08
5.......9/22/08.......9/28/08
6.......9/29/08.......10/5/08
7.......10/06/08.....10/12/08
8.......10/13/08......None
How do I get the week #, start date, end date for a given date?
I've always found it easiest and most efficient (for SQL Server) to construct a table with one row for every week into the future through your domain horizon; and join to that (with a "WHERE GETDATE() >= MONDATE AND NOT EXISTS (SELECT 1 FROM table WHERE MONDATE < GETDATE())".
Anything you try to do with UDF's will be much less efficient and I find more difficult to use.
You can get Monday for any given date in a week as:
DATEADD(d, 1 - DATEPART(dw, #date), #date)
and you can write a stored procedure with the following body
-- find Monday at that week
DECLARE #currentDate SMALLDATETIME
SELECT #currentDate = DATEADD(d, 1 - DATEPART(dw, #date), #date)
-- create a table and insert the first record
DECLARE #weekTable TABLE (Id INT, StartDate SMALLDATETIME, EndDate SMALLDATETIME)
INSERT INTO #weekTable VALUES (0, NULL, #currentDate)
-- increment the date
SELECT #currentDate = DATEADD(d, 1, #currentDate)
-- iterate for 7 more weeks
DECLARE #id INT
SET #id = 1
WHILE #id < 8
BEGIN
INSERT INTO #weekTable VALUES (#id, #currentDate, DATEADD(d, 6, #currentDate))
SELECT #currentDate = DATEADD(ww, 1, #currentDate)
SET #id = #id + 1
END
-- add the last record
INSERT INTO #weekTable VALUES (8, #currentDate, NULL)
-- select the values
SELECT Id 'Week #', StartDate 'Start Date', EndDate 'End Date'
FROM #weekTable
When I pass
#date = '20080827'
to this procedure, I get the following
Week # Start Date End Date
0 NULL 2008-08-24 00:00:00
1 2008-08-25 00:00:00 2008-08-31 00:00:00
2 2008-09-01 00:00:00 2008-09-07 00:00:00
3 2008-09-08 00:00:00 2008-09-14 00:00:00
4 2008-09-15 00:00:00 2008-09-21 00:00:00
5 2008-09-22 00:00:00 2008-09-28 00:00:00
6 2008-09-29 00:00:00 2008-10-05 00:00:00
7 2008-10-06 00:00:00 2008-10-12 00:00:00
8 2008-10-13 00:00:00 NULL
--SQL sets the first day of the week as sunday and for our purposes we want it to be Monday.
--This command does that.
SET DATEFIRST 1
DECLARE
#ReportDate DATETIME,
#Weekday INTEGER,
#NumDaysToMonday INTEGER,
#MondayStartPoint DATETIME,
#MondayStartPointWeek INTEGER,
#DateToProcess DATETIME,
#DateToProcessWeek INTEGER,
#Bucket VARCHAR(50),
#DaysDifference INTEGER,
#BucketNumber INTEGER,
#NumDaysToMondayOfDateToProcess INTEGER,
#WeekdayOfDateToProcess INTEGER,
#MondayOfDateToProcess DATETIME,
#SundayOfDateToProcess DATETIME
SET #ReportDate = '2009-01-01'
print #ReportDate
SET #DateToProcess = '2009-01-26'
--print #DateToProcess
SET #Weekday = (select DATEPART ( dw , #ReportDate ))
--print #Weekday
--print DATENAME(dw, #ReportDate)
SET #NumDaysToMonday =
(SELECT
CASE
WHEN #Weekday = 1 THEN 0
WHEN #Weekday = 2 THEN 1
WHEN #Weekday = 3 THEN 2
WHEN #Weekday = 4 THEN 3
WHEN #Weekday = 5 THEN 4
WHEN #Weekday = 6 THEN 5
WHEN #Weekday = 7 THEN 6
END)
--print #NumDaysToMonday
SET #MondayStartPoint = (SELECT DATEADD (d , -1*#NumDaysToMonday, #ReportDate))
--print #MondayStartPoint
SET #DaysDifference = DATEDIFF ( dd , #MondayStartPoint , #DateToProcess )
--PRINT #DaysDifference
SET #BucketNumber = #DaysDifference/7
--print #BucketNumber
----Calculate the start and end dates of this bucket------
PRINT 'Start Of New Calc'
print #DateToProcess
SET #WeekdayOfDateToProcess = (select DATEPART ( dw , #DateToProcess ))
print #WeekdayOfDateToProcess
SET #NumDaysToMondayOfDateToProcess=
(SELECT
CASE
WHEN #WeekdayOfDateToProcess = 1 THEN 0
WHEN #WeekdayOfDateToProcess = 2 THEN 1
WHEN #WeekdayOfDateToProcess = 3 THEN 2
WHEN #WeekdayOfDateToProcess = 4 THEN 3
WHEN #WeekdayOfDateToProcess = 5 THEN 4
WHEN #WeekdayOfDateToProcess = 6 THEN 5
WHEN #WeekdayOfDateToProcess = 7 THEN 6
END)
print #NumDaysToMondayOfDateToProcess
SET #MondayOfDateToProcess = (SELECT DATEADD (d , -1*#NumDaysToMondayOfDateToProcess, #DateToProcess))
print #MondayOfDateToProcess ---This is the start week
SET #SundayOfDateToProcess = (SELECT DATEADD (d , 6, #MondayOfDateToProcess))
PRINT #SundayOfDateToProcess
The problem I see with the one bucket at a time approach is that its hard to make it scale,
If you join into a user defined function you will get better performance, you could use this a a starting point
Why not use a combination of DATEPART(year, date-column) and DATEPART(week, date-column) and group by these values. This works if the week in DATEPART is aligned on Mondays as ISO 8601 requires. In outline:
SELECT DATEPART(year, date_column) AS yyyy,
DATEPART(week, date_column) AS ww,
...other material as required...
FROM SomeTableOrOther
WHERE ...appropriate filters...
GROUP BY yyyy, ww -- ...and other columns as necessary...