datetime difference and display in years

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datetime difference and display in years

scottichrosaviakosmos

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June 29, 2010 at 7:04 am

#236985

i want to get date difference like a start date and end date and find difference in years but i want like if for example one date is 2007 and end date is 1 jan 2010 then the difference is of 3 years and one day so sql is showing 3 year .. but i want it to show 4 years. i.e even one day is extra should add that also.

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ChrisM@Work SSC Guru Points: 186127 More actions · Answer 1

scottichrosaviakosmos (6/29/2010)
i want to get date difference like a start date and end date and find difference in years but i want like if for example one date is 2007 and end date is 1 jan 2010 then the difference is of 3 years and one day so sql is showing 3 year .. but i want it to show 4 years. i.e even one day is extra should add that also.

The year count increments with year boundaries;

SELECT DATEDIFF(dd, '31-12-2009', '01-01-2010') -- 1 day

SELECT DATEDIFF(yy, '31-12-2009', '01-01-2010') -- 1 year

It might be more useful for you to count months instead:

set dateformat dmy

SELECT DATEDIFF(mm, '31-12-2006', '31-12-2009') -- 36

SELECT DATEDIFF(mm, '31-12-2006', '01-01-2010') -- 37

SELECT DATEDIFF(mm, '01-01-2009', '31-12-2009') -- 11

SELECT DATEDIFF(mm, '01-11-2009', '31-01-2010') -- 2

^{“Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw}

For fast, accurate and documented assistance in answering your questions, please read this article.
Understanding and using APPLY, (I) and (II) Paul White
Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff Moden

scott.pletcher SSCertifiable Points: 5958 More actions · Answer 2

Add 1 to the DATEDIFF() unless the dates are equivalent.

SELECT DATEDIFF(YEAR, date1, date2) + CASE WHEN CONVERT(CHAR(5), date1, 1) = CONVERT(CHAR(5), date2, 1) THEN 0 ELSE 1 END

You have to decide if/how you want to handle Feb 28 vs Feb 29.

For example:

+ CASE WHEN CONVERT(CHAR(5), DATEADD(DAY, 1, date1), 1) = CONVERT(CHAR(5), DATEADD(DAY, 1, date2), 1) THEN 0 ELSE 1 END

would consider Feb 28 and Feb 29 as the "same day" for your calculation purposes.

Scott Pletcher, SQL Server MVP 2008-2010

Ron McCullough SSC Guru Points: 63877 More actions · Answer 3

scott.pletcher

CREATE TABLE #T(Id INT,Date1 DATETIME,Date2 DATETIME)

INSERT INTO #T

SELECT 1,'2/1/2007','1/31/2010' UNION ALL

SELECT 2,'2/1/2007','2/1/2010' UNION ALL

SELECT 3, '2/1/2007','1/15/2010'

SELECT Id, Date1 AS 'Start', Date2 AS 'End', DATEDIFF(YEAR, date1, date2)

+ CASE WHEN CONVERT(CHAR(5), date1, 1) = CONVERT(CHAR(5), date2, 1)

THEN 0 ELSE 1 END Years

FROM #T

Yields:

IdStart End Years

12007-02-01 00:00:00.0002010-01-31 00:00:00.0004

22007-02-01 00:00:00.0002010-02-01 00:00:00.0003

32007-02-01 00:00:00.0002010-01-15 00:00:00.0004

I think what you meant to write:

SELECT Id, Date1 AS 'Start', Date2 AS 'End', DATEDIFF(YEAR, date1, date2)

+ CASE WHEN CONVERT(CHAR(5), date1, 1) = CONVERT(CHAR(5), date2, 1)

THEN 1 ELSE 0 END Years

FROM #T

IdStart End Years

12007-02-01 00:00:00.0002010-01-31 00:00:00.0003

22007-02-01 00:00:00.0002010-02-01 00:00:00.0004

32007-02-01 00:00:00.0002010-01-15 00:00:00.0003

If everything seems to be going well, you have obviously overlooked something.

Ron

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scott.pletcher SSCertifiable Points: 5958 More actions · Answer 4

Actually, no, if the result was an exact number of years, I kinda' figured he wouldn't add one to that. But that was just my own best guess, of course 🙂

Scott Pletcher, SQL Server MVP 2008-2010

Lynn Pettis SSC Guru Points: 442467 More actions · Answer 5

Or is it more like this:

CREATE TABLE #T(Id INT,date1 DATETIME,date2 DATETIME)

INSERT INTO #T

SELECT 1,'2/1/2007','1/31/2010' UNION ALL

SELECT 2,'2/1/2007','2/1/2010' UNION ALL

SELECT 3, '2/1/2007','1/15/2010' UNION ALL

SELECT 4, '2/1/2007','2/2/2010'

SELECT

Id,

date1 AS 'Start',

date2 AS 'End',

DATEDIFF(YEAR, date1, date2) + CASE WHEN DATEADD(yy, -1 * DATEDIFF(YEAR, date1, date2), date2) > date1

THEN 1 ELSE 0

END Years

FROM #T

DROP TABLE #T

Jeff Moden SSC Guru Points: 1003851 More actions · Answer 6

I forget who I got this one from but I've never found anything simpler for correctly calculating age in years. Using Lynn's good test harness...

CREATE TABLE #T(Id INT,date1 DATETIME,date2 DATETIME)

INSERT INTO #T

SELECT 1,'2/1/2007','1/31/2010' UNION ALL

SELECT 2,'2/1/2007','2/1/2010' UNION ALL

SELECT 3, '2/1/2007','1/15/2010' UNION ALL

SELECT 4, '2/1/2007','2/2/2010'

SELECT

Id,

date1 AS 'Start',

date2 AS 'End',

YEAR(date2 - DATEPART(dy, date1) + 1) - YEAR(date1)

FROM #T

DROP TABLE #T

--Jeff Moden

RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.

Change is inevitable... Change for the better is not.

Helpful Links:
How to post code problems
How to Post Performance Problems
Create a Tally Function (fnTally)

Lynn Pettis SSC Guru Points: 442467 More actions · Answer 7

What would really help is for the OP to come back and tell us what is really needed or if any of the code snippets provided actually solve the problem.

Jeff Moden SSC Guru Points: 1003851 More actions · Answer 8

Agreed.

--Jeff Moden

RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.

Change is inevitable... Change for the better is not.

Helpful Links:
How to post code problems
How to Post Performance Problems
Create a Tally Function (fnTally)