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I need to extraxt a list of users who turn 15 years old in any day in a certain month (e.g. in June or in July) and I am trying to use string comparison in SQL Server (stored procedure) but it is not working.
I get #Month parameter from a SSRS which is given from a dropdown list of next 10 months. (some of this month will be in the next year).
I have an Age function which converts a date format from 27/07/2003 (BirthDate) to a string 15 years,2 months,27 days. Naturally, There are people aged 9 years, 0 months, 2 days as well.
So far, I could write the code to check if the person will be 15 and at least 1 month (15 years,1 months) in next July (#Month + 1) supposing we want to know if they turn 15 in June (#Month) but still it is not working because of string comparison.
Age(BirthDate, GETDATE()) is the function which turns age in this format:
15 years,2 months,27 days as string.
I hope it is clear what I mean.
declare #Age varchar (20) = 15,
#Month varchar (25) = 'June',
#CurrentMonth varchar (20) = null,
SET #CurrentMonth = DATENAME(month, GETDATE()); /* returns current month in string */
SELECT My_ID
,Title
,FirstName
,LastName
,Gender
,CONVERT(VARCHAR, BirthDate, 103) AS BirthDate
,dbo.Age(BirthDate, GETDATE()) AS Age
,LocalityName AS Locality
,GETDATE() AS ReportDate
,MONTH(GETDATE()) AS MONTH
,YEAR(GETDATE()) AS YEAR
FROM dbo.vw_individuals
WHERE (LEFT(dbo.Age(BirthDate, DATEADD(month, 1 +
(SELECT DATEDIFF(MONTH, #CurrentMonth + ' 01 2010', #Month + ' 01 2010')
+ CASE WHEN DATEDIFF(MONTH, #CurrentMonth + ' 01 2010', #Month + ' 01 2010') < 0 THEN 12 ELSE 0 END)
, GETDATE())), 2) = #Age)
AND (LEFT(dbo.Age(BirthDate, DATEADD(month, 1 +
(SELECT DATEDIFF(MONTH, #CurrentMonth + ' 01 2010', #Month + ' 01 2010')
+ CASE WHEN DATEDIFF(MONTH, #CurrentMonth + ' 01 2010', #Month + ' 01 2010') < 0 THEN 12 ELSE 0 END)
, GETDATE())), 16) < #Age + ' years,2 month')
AND (month(convert(DATETIME, BirthDate, 103)) = (SELECT DATEPART(MM, #Month + '01 2010'))) /* Checks if the BirthDate month is the same as the chosen month (#Month) */
so given the #Month and #Year you're interested in, then the people you need are given BY
SELECT ..... WHERE YEAR(BirthDate) = #YEAR - 15 AND Month(BirthDate) = #Month
so all the people returned have a 15th birthday in #Month/#Year
-- bit of a pain though, might not quite work if a person is born 29th February, they turn 15 in March although that will never be a leap year, so you could get away with
SELECT ..... WHERE YEAR(BirthDate) = #YEAR - 15
AND Month(BirthDate)
+ CASE WHEN MONTH(Birthdate) = 2 AND DAY(Birthdate) = 29 THEN 1 ELSE 0 END
= #Month
Below should show you way to go, your query is much too complicated for what you need (I think), just use logic from my IF in your WHERE
declare #dob datetime = '2003-10-13 17:21:45.620'
if(month(#dob) = month(getdate())
AND year(#dob)+15 = year(getdate()))
print '15 years old this month'
else
print 'not 15 years old this month'
The same yoke, just used table variable:
declare #dob datetime = '2003-10-13 17:21:45.620'
declare #t table (dob datetime)
insert into #t
values ('2003-10-13 17:21:45.620'), ('2004-10-13 17:21:45.620'), ('2003-10-30 17:21:45.620')
select *
from #t
where month(dob) = month(getdate())
and year(dob)+15 = year(getdate())
EDIT
Other way to achieve what you need is to use DATEDIFF
declare #dob datetime = '2003-10-13 17:21:45.620'
declare #t table (dob datetime)
insert into #t
values ('2003-10-13 17:21:45.620'),
('2004-10-13 17:21:45.620'),
('2003-10-30 17:21:45.620')
select *
from #t
where datediff(month, dob, getdate()) = 180 --(15 years * 12months)
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'm looking for a way to calculate the age of a person. I'm making a function for it that I can use later as well as this time.
Function should take a ID-Number which are 11 characters in sweden format which is 560404-1234.
And return the age of the person just the years. So far I've thought of making a subtraction with example: Getdate year 2014 - 19+IDNr datepart yy to get the sum.
Still stuck though.
Cheers
I am guessing that the Swedish id number format is YYMMDD followed by something else.
If so, then the following should work to get the date:
select cast('19'+left(idn, 6) as date)
Then the age would be:
select datediff(year, cast('19'+left(idn, 6) as date), getdate())
What do the Swedes do about people born in the last 14 or so years? What happens to those who are over 100 years old?
If "dean" is correct, then you can try following code,
Create FUNCTION [dbo].[udfn_Get_Age_In_Years]
(
#ID_Number varchar(50)
)
Returns int
--select dbo.[udfn_Get_Age_In_Years]('560404-1234')
as
begin
Declare #retVal int = 0
select #retVal = Datediff(mm,col_dob,GETDATE()) / 12 from myTable
where ID_Number = #ID_Number
Return #retVal
end
This is my answer it's similar to the above ones but this is how I've solved it after a while.
ALTER function [dbo].[Function_AGE]
(
#IdNr varchar(13)
)
returns int
as
begin
Declare #Calc int
Set #IdNr = (Select Case when LEN(#IdNr) > 11
then cast(left(#IdNr, 8) as date)
else cast('19'+left(#IdNr,6) as date) end)
set #Calc = (select datediff(year,#IdNr,getdate()))
Return #Calc
end
Here is how you would calculate age of a person, given their age 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
To Compare a birthdate to #Date, the DATEDIFF gets you close. Subtract off of that if the month and day haven't occurred yet in the year.
CASE WHEN DATEPART(mm,birthdate) > DATEPART(mm,#Date)
THEN DATEDIFF(YEAR, birthdate AS DATE), #Date) - 1
WHEN DATEPART(mm,birthdate) = DATEPART(mm,#Date)
AND DATEPART(dd,birthdate) > DATEPART(dd,#Date)
THEN DATEDIFF(YEAR, birthdate AS DATE), #Date) - 1
ELSE DATEDIFF(YEAR, CAST(birthdate AS DATE), #Date)
END AS Age
This is harder than it looks. I need a function that calculates the numbers of a given weekday in a date range. I don't want any loops or recursive SQL. There are millions of examples doing just that. I need a fast function for calculation.
Input of the function will be weekday, fromdata, todate
-- counting fridays
set datefirst 1
SELECT dbo.f_countweekdays(5, '2011-07-01', '2011-07-31'),
dbo.f_countweekdays(5, '2011-07-08', '2011-07-15'),
dbo.f_countweekdays(5, '2011-07-09', '2011-07-15'),
dbo.f_countweekdays(5, '2011-07-09', '2011-07-14')
Expected result:
5, 2, 1, 0
create function dbo.f_countweekdays
(
#DOW int,
#StartDate datetime,
#EndDate datetime
)
returns int
begin
return
( select datediff(wk, T2.St, T2.En) -
case when T1.SDOW > #DOW then 1 else 0 end -
case when T1.EDOW < #DOW then 1 else 0 end
from (select datepart(dw, #StartDate),
datepart(dw, #EndDate)) as T1(SDOW, EDOW)
cross apply (select dateadd(d, - T1.SDOW, #StartDate),
dateadd(d, 7 - T1.EDOW, #EndDate)) as T2(St, En))
end
#Mikael Eriksson has got a wonderful idea, but his implementation seems a bit overcomplicated.
Here's what I've come up with (and I'd like to stress that it is based on the solution by #Mikael, to whom the main credit should go):
ALTER FUNCTION dbo.f_countweekdays (#Dow int, #StartDate datetime, #EndDate datetime)
RETURNS int
AS BEGIN
RETURN (
SELECT
DATEDIFF(wk, #StartDate, #EndDate)
- CASE WHEN DATEPART(dw, #StartDate) > #Dow THEN 1 ELSE 0 END
- CASE WHEN DATEPART(dw, #EndDate) < #Dow THEN 1 ELSE 0 END
+ 1
)
END
UPDATE
As Mikael has correctly noted in his answer's comment thread, in order for the above solution to work correctly the DATEFIRST setting must be set to 7 (Sunday). Although I couldn't find this documented, a quick test revealed that DATEDIFF(wk) disregards the actual DATEFIRST setting and indeed returns the difference in weeks as if DATEFIRST was always set to 7. At the same time DATEPART(dw) does respect DATEFIRST, so with DATEFIRST set to a value other than 7 the two functions return mutually inconsistent results.
Therefore, the above script must be amended in order to account for different values of the DATEFIRST setting when calculating DATEDIFF(wk). Happily, the fix doesn't seem to have made the solution much more complicated than before, in my opinion. Judge for yourself, though:
ALTER FUNCTION dbo.f_countweekdays (#Dow int, #StartDate datetime, #EndDate datetime)
RETURNS int
AS BEGIN
RETURN (
SELECT
DATEDIFF(wk, DATEADD(DAY, -##DATEFIRST, #StartDate),
DATEADD(DAY, -##DATEFIRST, #EndDate))
- CASE WHEN DATEPART(dw, #StartDate) > #Dow THEN 1 ELSE 0 END
- CASE WHEN DATEPART(dw, #EndDate) < #Dow THEN 1 ELSE 0 END
+ 1
)
END
Edited: both -##DATEFIRST % 7 entries have been simplified to just -##DATEFIRST, as someone suggested here.
An alternative approach is the good old-fashioned data warehouse time dimension, where you have a table with all potential dates in it, along with any useful information you want to filter/count by:
Key ActualDate DayName IsWeekday DayNumberInYear FinancialQuarter
20110101 1 Jan 2011 Saturday 0 1 2011 Q1
20110102 2 Jan 2011 Sunday 0 2 2011 Q1
20110103 3 Jan 2011 Monday 1 3 2011 Q1
Then just join to that table and filter, e.g.
SELECT
COUNT(*)
FROM
date_dimension
WHERE
ActualDate BETWEEN '1 Jan 2011' AND '3 Jan 2011' AND
IsWeekday = 1
If you do date analysis a lot over a known range of dates, this can really speed up and simplify your queries. Whether you know your possible date ranges in advance is the limiting factor on whether this is helpful, really, but it's a useful trick to know about.
I need to determine the number of days in a month for a given date in SQL Server.
Is there a built-in function? If not, what should I use as the user-defined function?
In SQL Server 2012 you can use EOMONTH (Transact-SQL) to get the last day of the month and then you can use DAY (Transact-SQL) to get the number of days in the month.
DECLARE #ADate DATETIME
SET #ADate = GETDATE()
SELECT DAY(EOMONTH(#ADate)) AS DaysInMonth
You can use the following with the first day of the specified month:
datediff(day, #date, dateadd(month, 1, #date))
To make it work for every date:
datediff(day, dateadd(day, 1-day(#date), #date),
dateadd(month, 1, dateadd(day, 1-day(#date), #date)))
Most elegant solution: works for any #DATE
DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,#DATE),0)))
Throw it in a function or just use it inline. This answers the original question without all the extra junk in the other answers.
examples for dates from other answers:
SELECT DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,'1/31/2009'),0))) Returns 31
SELECT DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,'2404-feb-15'),0))) Returns 29
SELECT DAY(DATEADD(DD,-1,DATEADD(MM,DATEDIFF(MM,-1,'2011-12-22'),0))) Returns 31
--Last Day of Previous Month
SELECT DATEPART(day, DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE()),0)))
--Last Day of Current Month
SELECT DATEPART(day, DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE())+1,0)))
--Last Day of Next Month
SELECT DATEPART(day, DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,GETDATE())+2,0)))
Personally though, I would make a UDF for it if there is not a built in function...
I would suggest:
SELECT DAY(EOMONTH(GETDATE()))
This code gets you the number of days in current month:
SELECT datediff(dd,getdate(),dateadd(mm,1,getdate())) as datas
Change getdate() to the date you need to count days for.
--- sql server below 2012---
select day( dateadd(day,-1,dateadd(month, 1, convert(date,'2019-03-01'))))
-- this for sql server 2012--
select day(EOMONTH(getdate()))
Solution 1: Find the number of days in whatever month we're currently in
DECLARE #dt datetime
SET #dt = getdate()
SELECT #dt AS [DateTime],
DAY(DATEADD(mm, DATEDIFF(mm, -1, #dt), -1)) AS [Days in Month]
Solution 2: Find the number of days in a given month-year combo
DECLARE #y int, #m int
SET #y = 2012
SET #m = 2
SELECT #y AS [Year],
#m AS [Month],
DATEDIFF(DAY,
DATEADD(DAY, 0, DATEADD(m, ((#y - 1900) * 12) + #m - 1, 0)),
DATEADD(DAY, 0, DATEADD(m, ((#y - 1900) * 12) + #m, 0))
) AS [Days in Month]
You do need to add a function, but it's a simple one. I use this:
CREATE FUNCTION [dbo].[ufn_GetDaysInMonth] ( #pDate DATETIME )
RETURNS INT
AS
BEGIN
SET #pDate = CONVERT(VARCHAR(10), #pDate, 101)
SET #pDate = #pDate - DAY(#pDate) + 1
RETURN DATEDIFF(DD, #pDate, DATEADD(MM, 1, #pDate))
END
GO
SELECT Datediff(day,
(Convert(DateTime,Convert(varchar(2),Month(getdate()))+'/01/'+Convert(varchar(4),Year(getdate())))),
(Convert(DateTime,Convert(varchar(2),Month(getdate())+1)+'/01/'+Convert(varchar(4),Year(getdate()))))) as [No.of Days in a Month]
select datediff(day,
dateadd(day, 0, dateadd(month, ((2013 - 1900) * 12) + 3 - 1, 0)),
dateadd(day, 0, dateadd(month, ((2013 - 1900) * 12) + 3, 0))
)
Nice Simple and does not require creating any functions Work Fine
You need to create a function, but it is for your own convenience. It works perfect and I never encountered any faulty computations using this function.
CREATE FUNCTION [dbo].[get_days](#date datetime)
RETURNS int
AS
BEGIN
SET #date = DATEADD(MONTH, 1, #date)
DECLARE #result int = (select DAY(DATEADD(DAY, -DAY(#date), #date)))
RETURN #result
END
How it works: subtracting the date's day number from the date itself gives you the last day of previous month. So, you need to add one month to the given date, subtract the day number and get the day component of the result.
select add_months(trunc(sysdate,'MM'),1) - trunc(sysdate,'MM') from dual;
I upvoted Mehrdad, but this works as well. :)
CREATE function dbo.IsLeapYear
(
#TestYear int
)
RETURNS bit
AS
BEGIN
declare #Result bit
set #Result =
cast(
case when ((#TestYear % 4 = 0) and (#testYear % 100 != 0)) or (#TestYear % 400 = 0)
then 1
else 0
end
as bit )
return #Result
END
GO
CREATE FUNCTION dbo.GetDaysInMonth
(
#TestDT datetime
)
RETURNS INT
AS
BEGIN
DECLARE #Result int
DECLARE #MonthNo int
Set #MonthNo = datepart(m,#TestDT)
Set #Result =
case #MonthNo
when 1 then 31
when 2 then
case
when dbo.IsLeapYear(datepart(yyyy,#TestDT)) = 0
then 28
else 29
end
when 3 then 31
when 4 then 30
when 5 then 31
when 6 then 30
when 7 then 31
when 8 then 31
when 9 then 30
when 10 then 31
when 11 then 30
when 12 then 31
end
RETURN #Result
END
GO
To Test
declare #testDT datetime;
set #testDT = '2404-feb-15';
select dbo.GetDaysInMonth(#testDT)
here's another one...
Select Day(DateAdd(day, -Day(DateAdd(month, 1, getdate())),
DateAdd(month, 1, getdate())))
I know this question is old but I thought I would share what I'm using.
DECLARE #date date = '2011-12-22'
/* FindFirstDayOfMonth - Find the first date of any month */
-- Replace the day part with -01
DECLARE #firstDayOfMonth date = CAST( CAST(YEAR(#date) AS varchar(4)) + '-' +
CAST(MONTH(#date) AS varchar(2)) + '-01' AS date)
SELECT #firstDayOfMonth
and
DECLARE #date date = '2011-12-22'
/* FindLastDayOfMonth - Find what is the last day of a month - Leap year is handled by DATEADD */
-- Get the first day of next month and remove a day from it using DATEADD
DECLARE #lastDayOfMonth date = CAST( DATEADD(dd, -1, DATEADD(mm, 1, FindFirstDayOfMonth(#date))) AS date)
SELECT #lastDayOfMonth
Those could be combine to create a single function to retrieve the number of days in a month if needed.
SELECT DAY(SUBDATE(ADDDATE(CONCAT(YEAR(NOW()), '-', MONTH(NOW()), '-1'), INTERVAL 1 MONTH), INTERVAL 1 DAY))
Nice 'n' Simple and does not require creating any functions
Mehrdad Afshari reply is most accurate one, apart from usual this answer is based on formal mathematical approach given by Curtis McEnroe in his blog https://cmcenroe.me/2014/12/05/days-in-month-formula.html
DECLARE #date DATE= '2015-02-01'
DECLARE #monthNumber TINYINT
DECLARE #dayCount TINYINT
SET #monthNumber = DATEPART(MONTH,#date )
SET #dayCount = 28 + (#monthNumber + floor(#monthNumber/8)) % 2 + 2 % #monthNumber + 2 * floor(1/#monthNumber)
SELECT #dayCount + CASE WHEN #dayCount = 28 AND DATEPART(YEAR,#date)%4 =0 THEN 1 ELSE 0 END -- leap year adjustment
To get the no. of days in a month we can directly use Day() available in SQL.
Follow the link posted at the end of my answer for SQL Server 2005 / 2008.
The following example and the result are from SQL 2012
alter function dbo.[daysinm]
(
#dates nvarchar(12)
)
returns int
as
begin
Declare #dates2 nvarchar(12)
Declare #days int
begin
select #dates2 = (select DAY(EOMONTH(convert(datetime,#dates,103))))
set #days = convert(int,#dates2)
end
return #days
end
--select dbo.daysinm('08/12/2016')
Result in SQL Server SSMS
(no column name)
1 31
Process:
When EOMONTH is used, whichever the date format we use it is converted into DateTime format of SQL-server. Then the date output of EOMONTH() will be 2016-12-31 having 2016 as Year, 12 as Month and 31 as Days.
This output when passed into Day() it gives you the total days count in the month.
If we want to get the instant result for checking we can directly run the below code,
select DAY(EOMONTH(convert(datetime,'08/12/2016',103)))
or
select DAY(EOMONTH(convert(datetime,getdate(),103)))
for reference to work in SQL Server 2005/2008/2012, please follow the following external link ...
Find No. of Days in a Month in SQL
DECLARE #date DATETIME = GETDATE(); --or '12/1/2018' (month/day/year)
SELECT DAY(EOMONTH ( #date )) AS 'This Month';
SELECT DAY(EOMONTH ( #date, 1 )) AS 'Next Month';
result:
This Month
31
Next Month
30
DECLARE #m int
SET #m = 2
SELECT
#m AS [Month],
DATEDIFF(DAY,
DATEADD(DAY, 0, DATEADD(m, +#m -1, 0)),
DATEADD(DAY, 0, DATEADD(m,+ #m, 0))
) AS [Days in Month]
RETURN day(dateadd(month, 12 * #year + #month - 22800, -1))
select day(dateadd(month, 12 * year(date) + month(date) - 22800, -1))
A cleaner way of implementing this is using the datefromparts function to construct the first day of the month, and calculate the days from there.
CREATE FUNCTION [dbo].[fn_DaysInMonth]
(
#year INT,
#month INT
)
RETURNS INT
AS
BEGIN
IF #month < 1 OR #month > 12 RETURN NULL;
IF #year < 1753 OR #year > 9998 RETURN NULL;
DECLARE #firstDay DATE = datefromparts(#year, #month, 1);
DECLARE #lastDay DATE = dateadd(month, 1, #firstDay);
RETURN datediff(day, #firstDay, #lastDay);
END
GO
Similarily, you can calculate the days in a year:
CREATE FUNCTION [dbo].[fn_DaysInYear]
(
#year INT
)
RETURNS INT
AS
BEGIN
IF #year < 1753 OR #year > 9998 RETURN NULL;
DECLARE #firstDay DATE = datefromparts(#year, 1, 1);
DECLARE #lastDay DATE = dateadd(year, 1, #firstDay);
RETURN datediff(day, #firstDay, #lastDay);
END
GO
use SQL Server EOMONTH Function nested with day to get last day of month
select Day(EOMONTH('2020-02-1')) -- Leap Year returns 29
select Day(EOMONTH('2021-02-1')) -- returns 28
select Day(EOMONTH('2021-03-1')) -- returns 31
For any date
select DateDiff(Day,#date,DateAdd(month,1,#date))
select first_day=dateadd(dd,-1*datepart(dd,getdate())+1,getdate()),
last_day=dateadd(dd,-1*datepart(dd,dateadd(mm,1,getdate())),dateadd(mm,1,getdate())),
no_of_days = 1+datediff(dd,dateadd(dd,-1*datepart(dd,getdate())+1,getdate()),dateadd(dd,-1*datepart(dd,dateadd(mm,1,getdate())),dateadd(mm,1,getdate())))
replace any date with getdate to get the no of months in that particular date
DECLARE #Month INT=2,
#Year INT=1989
DECLARE #date DateTime=null
SET #date=CAST(CAST(#Year AS nvarchar) + '-' + CAST(#Month AS nvarchar) + '-' + '1' AS DATETIME);
DECLARE #noofDays TINYINT
DECLARE #CountForDate TINYINT
SET #noofDays = DATEPART(MONTH,#date )
SET #CountForDate = 28 + (#noofDays + floor(#noofDays/8)) % 2 + 2 % #noofDays + 2 * floor(1/#noofDays)
SET #noofDays= #CountForDate + CASE WHEN #CountForDate = 28 AND DATEPART(YEAR,#date)%4 =0 THEN 1 ELSE 0 END
PRINT #noofDays
DECLARE #date nvarchar(20)
SET #date ='2012-02-09 00:00:00'
SELECT DATEDIFF(day,cast(replace(cast(YEAR(#date) as char)+'-'+cast(MONTH(#date) as char)+'-01',' ','')+' 00:00:00' as datetime),dateadd(month,1,cast(replace(cast(YEAR(#date) as char)+'-'+cast(MONTH(#date) as char)+'-01',' ','')+' 00:00:00' as datetime)))
simple query in SQLServer2012 :
select day(('20-05-1951 22:00:00'))
i tested for many dates and it return always a correct result