May 10, 2009 at 10:53 pm
Here I have used some queries but i unable to get a exact solution...
Please try else Am I Correct?
My quires:
SELECT EMP_CODE,DATE_OF_BIRTH ,
DATEDIFF(DD,DATE_OF_BIRTH,GETDATE())/365 AS AGE
FROM EMP_MST
May 11, 2009 at 12:27 am
Hi,
Your output gives only the days,
The age may come in the format of years months and days like 34 years 10Months and 15 days
ARUN SAS
May 11, 2009 at 12:32 am
Ok thanks for replay...
Tell me a another way to fetch the exact age of Different Date...
May 11, 2009 at 1:16 am
Hi,
you can try this
select EMP_CODE,DATE_OF_BIRTH ,DATEDIFF(DD,DATE_OF_BIRTH,getdate())/365 as years,(DATEDIFF(DD,DATE_OF_BIRTH,getdate())%365)/30 as months,
DATEDIFF(DD,DATE_OF_BIRTH,getdate())%365%30 as days
from EMP_MST where accountid=4
Hitendra
May 11, 2009 at 1:16 am
Hi,
you can try this
select EMP_CODE,DATE_OF_BIRTH ,DATEDIFF(DD,DATE_OF_BIRTH,getdate())/365 as years,(DATEDIFF(DD,DATE_OF_BIRTH,getdate())%365)/30 as months,
DATEDIFF(DD,DATE_OF_BIRTH,getdate())%365%30 as days
from EMP_MST where accountid=4
Hitendra
May 11, 2009 at 1:16 am
Hi,
you can try this
select EMP_CODE,DATE_OF_BIRTH ,DATEDIFF(DD,DATE_OF_BIRTH,getdate())/365 as years,(DATEDIFF(DD,DATE_OF_BIRTH,getdate())%365)/30 as months,
DATEDIFF(DD,DATE_OF_BIRTH,getdate())%365%30 as days
from EMP_MST where accountid=4
Hitendra
May 11, 2009 at 2:11 am
t.hitendra (5/11/2009)
Hi,you can try this
select EMP_CODE,DATE_OF_BIRTH ,DATEDIFF(DD,DATE_OF_BIRTH,getdate())/365 as years,(DATEDIFF(DD,DATE_OF_BIRTH,getdate())%365)/30 as months,
DATEDIFF(DD,DATE_OF_BIRTH,getdate())%365%30 as days
from EMP_MST where accountid=4
Hitendra
Hi,
create the function like
CREATE FUNCTION AGE_ASOF_NOW (@EMP_DOB DATETIME,@curr_dt datetime)
returns varchar(1000)
AS
BEGIN
declare @AGE_ASOF_NOW varchar(100)
declare @DDIFF int
select @DDIFF = datediff(DAY,@EMP_DOB,@curr_dt)
select @AGE_ASOF_NOW = cast((@DDIFF/365)as varchar(4))+' Years '+ cast((@DDIFF%365)/30 as varchar(3))+' Months '+ cast((@DDIFF%365)%30 as varchar(3))+' Days '
return @AGE_ASOF_NOW
END
and call this function in the select statement like
select Dbo.AGE_ASOF_NOW (DATE_OF_BIRTH,getdate())
ARUN SAS
May 13, 2009 at 8:13 pm
Please be careful using ANY advice from an open forum such as this one without testing it first. The algorithms offered so far are severely lacking in accuracy. Although as schoolchildren, we learned that there are 365 days in a year, most of us would agree that because of the nature of the rules for leap years, we cannot use either 365 or 365.25 as an accurate divisor. And, of course, you don't believe that every month has 30 days.
I do appreciate the opportunity your request and the interim responses gave me to work up what I now believe to be an accurate script for returning an age in years, months and days. As has been suggested, it would probably be best implemented as a UDF (user defined function).
Oh. One more thing. Despite my confidence in this code, you must be sure of it before using it. Test it with as many weird combinations of @Today and @DOB as you can to prove out its accuracy.
Declare @DOB smallDatetime
,@Today smallDatetime
,@AgeYears int
,@AgePlusMonths int
,@AgePlusDays int
,@LastBD datetime
,@LastMonthBD datetime
--======================--
set @DoB = '20040228'
set @Today = '20080229'
--======================--
select
@AgeYears
= year(@today)-year(@dob)
- case when month(@today) < month(@dob)
or (month(@today) = month(@dob) and day(@today) < day(@dob))
then 1
else 0
end
,@LastBD = dateadd(year,@AgeYears,@Dob)
,@AgePlusMonths = datediff(month, @LastBD, @Today)
- case when month(@today) <= month(@LastBD) and day(@today) < day(@LastBD)
then 1
else 0
end
- case when month(@dob) = 2 and day(@dob) = 29
then 1
else 0
end
,@LastMonthBD = dateadd(month,@AgePlusMonths,@LastBD)
,@AgePlusDays = datediff(day, @LastMonthBD, @Today)
Select @today as today
,@DoB DoB
,@LastBD as LastBD
,@LastMonthBD as LastMonthBD
,@AgeYears AgeYears
,@AgePlusMonths As AgePlusMonths
,@AgePlusDays as AgePlusDays
May 14, 2009 at 2:12 am
declare @birthday datetime
declare @ToDate datetime
DECLARE @smonth TINYINT, @sday TINYINT, @syear SMALLINT
DECLARE @emonth TINYINT, @eday TINYINT, @eyear SMALLINT
DECLARE @months TINYINT, @days TINYINT, @years SMALLINT
DECLARE @tdate SMALLDATETIME
set @birthday = '15 feb 1969'
--set @retireage = 65
set @birthday = @birthday--getdate()
set @ToDate = getdate()--dateadd(year, +@retireage, @birthday)
--print @ToDate
SET @smonth = MONTH(@birthday)
SET @sday = DAY(@birthday)
SET @syear = YEAR(@birthday)
SET @emonth = MONTH(@ToDate)
SET @eday = DAY(@ToDate)
SET @eyear = YEAR(@ToDate)
SET @years = @eyear - @syear
SET @months = 0
SET @days = 0
IF (@emonth >= @smonth)
SET @months = @emonth - @smonth
ELSE
BEGIN
SET @years = @years - 1
SET @months = @emonth + 12 - @smonth
END
IF (@eday >= @sday)
SET @days = @eday - @sday
ELSE
BEGIN
IF (@months > 0)
SET @months = @months - 1
ELSE
BEGIN
SET @years = @years - 1
SET @months = @months + 11
END
SET @tdate = DATEADD(yy,@years,@birthday)
SET @tdate = DATEADD(m,@months,@tdate)
SET @days = DATEDIFF(d,@tdate,@ToDate)
END
print 'Accurate Age: (' +convert(varchar(11), @ToDate, 113) + ') ' +
convert(varchar(3), @years) + ' Years ' + convert(varchar(3), @months) + ' Months ' + convert(varchar(3), @days) + ' Days ' +
substring(CONVERT(VARCHAR(8),DATEADD(ss,datediff(second, @birthday, @todate),0),108), 1, 2) + ' Hours ' + substring(CONVERT(VARCHAR(8),DATEADD(ss,datediff(second, @birthday, @todate),0),108), 4, 2) + ' Minutes ' + substring(CONVERT(VARCHAR(8),DATEADD(ss,datediff(second, @birthday, @todate),0),108), 7, 2) + ' Seconds '
May 14, 2009 at 12:19 pm
+1 on testing with weird dates (both date of birth and as-of dates).
You must also know your business rule for leap day babies in non-leap-years. A version of what I use (not cleaned up! Please excuse bad table names, bad capitalization, and so on) follows.
Note there is a test query and small selection of interesting test cases that can be run as required in the comments at the beginning of the function.
create function dbo.ddfn_UT_AgeInYears
(
@START_DATEdatetime,
@END_DATEdatetime
)
returnsint
as
/*
Original code came from http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=74462
Original name F_AGE_IN_YEARS.
Updated 20080109 PH - comments updated.
Updated 20080423 PH - comments updated.
Function ddfn_UT_AgeInYears computes Age in years.
Input parameters @START_DATE and @END_DATE
are required. If either or both parameters
are null, the function returns null.
if @START_DATE midnight is greater than
@END_DATE midnight, the function returns NULL.
i.e. SELECT dbo.ddfn_UT_AgeInYears('1988-02-29','2007-02-28')
Age is defined as the number of anniversary dates
reached or passed from @START_DATE through @END_DATE.
Age is calculated based on midnight (00:00:00.000)
of parameters @START_DATE and @END_DATE.
Time of day is not used in the calculation.
NULL is returned when the @END_DATE is earlier
than the @START_DATE.
For example, someone born 2000-02-15
would be 5 years old on 2006-02-14,
but 6 years old on 2006-02-15.
Someone born on Feb 29 would be a year
older on Feb 28 in non-leap years, but
would be a year older on Feb 29 in leap years.
Function is valid for entire range of datetime
values from 1753-01-01 00:00:00.000 to
9999-12-31 23:59:59.997.
Quick test code:
declare @strSql varchar(8000)
set @strSql = ''
SET @strSql = 'select '
+ 'DOB AS BirthDate, '
+ 'AsOfDt AS AsOfDate, '
+ 'database.dbo.ddfn_UT_AgeInYears(DOB,AsOfDt) AS UT_AgeInYears'
+ ' FROM '
+ ' (SELECT ''2004-02-29'' AS DOB, ''2004-02-28'' AS AsOfDt'
+ ' union all SELECT ''2004-02-29'' AS DOB, ''2004-02-29'' AS AsOfDt'
+ ' union all SELECT ''2004-02-29'' AS DOB, ''2004-03-01'' AS AsOfDt'
+ ' union all SELECT '''' AS DOB, '''' AS AsOfDt'
+ ' union all SELECT ''1988-02-29'' AS DOB, ''2007-02-27'' AS AsOfDt'
+ ' union all SELECT ''1988-02-29'' AS DOB, ''2007-02-28'' AS AsOfDt'
+ ' union all SELECT ''1988-02-29'' AS DOB, ''2007-03-01'' AS AsOfDt'
+ ' union all SELECT '''' AS DOB, '''' AS AsOfDt'
+ ' union all SELECT ''1988-02-29'' AS DOB, ''2008-02-28'' AS AsOfDt'
+ ' union all SELECT ''1988-02-29'' AS DOB, ''2008-02-29'' AS AsOfDt'
+ ' union all SELECT ''1988-02-29'' AS DOB, ''2008-03-01'' AS AsOfDt'
+ ') HardCodeDateList'
exec (@strSQL)
*/
begin
declare @AGE_IN_YEARS int
-- Start out by setting the times to midnight.
select@START_DATE= dateadd(dd,datediff(dd,0,@START_DATE),0),
@END_DATE= dateadd(dd,datediff(dd,0,@END_DATE),0)
if @START_DATE > @END_DATE
begin
return null
end
select
@AGE_IN_YEARS =
datediff(yy,StartDateYearStart,EndDateYearStart) +
-- Subtract 1 if anniversary date is after end date
case
when AnniversaryThisYear <= @END_DATE
then 0
else -1
end
from
(
selectAnniversaryThisYear =
dateadd(yy,datediff(yy,StartDateYearStart,EndDateYearStart),@START_DATE),
StartDateYearStart,
EndDateYearStart
from
(
selectStartDateYearStart =
dateadd(yy,datediff(yy,0,@START_DATE),0),
EndDateYearStart =
dateadd(yy,datediff(yy,0,@END_DATE),0)
) aa
) a
return @AGE_IN_YEARS
end
May 14, 2009 at 1:07 pm
These are links to my original post of the function from the prior post on this thread, plus another that you might find useful.
Computing the age of someone is more difficult than it might seem when you take into account different month lengths, leap year, and other things.
This function returns age in format YYYY MM DD.
Age Function F_AGE_YYYY_MM_DD:
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=62729
This function returns age in years.
Age Function F_AGE_IN_YEARS:
May 14, 2009 at 1:16 pm
November 16, 2012 at 2:01 am
Simple way to compute your age accurately.
A lot of query not exactly compute the age? yes! because sometime they only compute the datediff between DOB and datenow and divide it to 365.25 days and as a result they get a number with decimal something like 25 is the age with .06 decimal (age=25.06).
In this query you exactly get the age as is it.
Example 1
DOB = 11/15/1987 and
datenow =11/15/2012 the result would be
AGE=25
Example 2
DOB = 11/14/1987 and
datenow =11/15/2012 the result would be
AGE=24
SO HERE ARE THE QUERY
DECLARE @DOB SMALLDATETIME
SELECT @DOB = '11/15/1987'
SELECT
CASE
WHEN MONTH(@DOB) >= MONTH(GETDATE()) AND DAY(@DOB) >=DAY(GETDATE()) THEN DATEDIFF(YY,@DOB,GETDATE())
ELSE DATEDIFF(YY,@DOB,GETDATE())-1
END AS AGE
HOPE I CAN HELP! 🙂
November 16, 2012 at 2:02 am
Simple way to compute your age accurately.
A lot of query not exactly compute the age? yes! because sometime they only compute the datediff between DOB and datenow and divide it to 365.25 days and as a result they get a number with decimal something like 25 is the age with .06 decimal (age=25.06).
In this query you exactly get the age as is it.
Example 1
DOB = 11/15/1987 and
datenow =11/15/2012 the result would be
AGE=25
Example 2
DOB = 11/14/1987 and
datenow =11/15/2012 the result would be
AGE=24
SO HERE ARE THE QUERY
DECLARE @DOB SMALLDATETIME
SELECT @DOB = '11/15/1987'
SELECT
CASE
WHEN MONTH(@DOB) >= MONTH(GETDATE()) AND DAY(@DOB) >=DAY(GETDATE()) THEN DATEDIFF(YY,@DOB,GETDATE())
ELSE DATEDIFF(YY,@DOB,GETDATE())-1
END AS AGE
HOPE I CAN HELP! 🙂
January 23, 2013 at 8:34 am
The following combinations give a negative result (that is fi 10 yr 3 months and -17 days):
(@dob, @today)
19270424, 19920707
19150131, 20000606
19230622, 20031215
19170124, 20020322
I cannot figure out why that is. Any ideas?
Grz,
Robert
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