feb 29 datediff error Ledyard Iowa

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feb 29 datediff error Ledyard, Iowa

I wrote the tip to highlight an issue -- namely, that the DATEDIFF function counts boundary crossing and not age in dateparts. SIGN() of the difference gives us 1, 0 or -1, all good except for > 1. For example, if you check for "y" it will say that there is 1 year between 12-31-1899 and 01-01-1900, as the difference between 1899 and 1900 is 1; months and days This is a VBA question not a VB6 question.

Thursday, July 18, 2013 - 8:50:31 PM - Rick Back To Top This works for me - the best of everything I could find. Tweet Thread Tools Show Printable Version Subscribe to this Thread… Display Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Apr 10th, 2008,09:17 AM #1 guosa View Profile View Forum which is probably OK. An issue can arise, as originally noted, when the year boundary is crossed but the DOY boundary has not so the age must be adjusted by -1.I think a big motivator

Calculating A Person's Age A frequent use of the DATEDIF is to compute someone's age based on the current date and their birthday. Not the answer you're looking for? See the start date of September 10, 2002. Cactus Data ApS, CPH. ' 2000-12-16.

I have removed the case statement but an argument could be made against my query being more elegant as it’s not as readable or clear as to what is happening. Or should it be DifferenceField Linq ;0)> Hope this helps! Rick Dobson, Thank you for this article. Because there is no boundary separating these two dates, the Int.

Adding Months And Years The DATEDIF Function Date Intervals Dates And Times Date And Time Entry Date And Time Arithmetic Distributing Dates Over Intervals Holidays Julian Dates Latitude And Longitude Overtime All dates are in US notation (mm/dd/yyyy). Notice that the 2nd column represents a leap year value - 2012-02-29. the author (me) didn't realize the implications of using DATEDIFF SECOND on precise timestamp values.

This will give -6 days if that's applicable to your dates: Declare @StartDate datetime, @EndDate datetime, @Result varchar(50) SET @StartDate = (Select '2013-07-19 12:20:07.097') Set @EndDate = (SELECT '2013-07-19 15:02:38.000') SET Given a firm grasp of this issue, you can engineer a solution to compute in age for any datepart that you require. A bullet shot into a door vs. Why are so many metros underground?

How to handle a senior developer diva who seems unaware that his skills are obsolete? throws a syntax error. i think it should be 0 month, 30 days With march don't work very well! :) php datetime intervals share|improve this question edited Dec 24 '12 at 0:28 hakre 134k26216387 asked Reply With Quote Quick Navigation Office Development Top Site Areas Settings Private Messages Subscriptions Who's Online Search Forums Forums Home Forums Visual Basic Visual Basic .NET VB.net CodeBank Visual Basic 6

AgeInYears value is 0 for both start dates. Solution The SQL Server online documentation describes how to compute the number of datepart boundary crossings between two dates and/or times with the DATEDIFF function. Interval indicates the unit of time that =DATEDIF is to return is result.

Date1 must be less than (earlier) or equal to Date2. If Day(DateSerial(Year(datDate1), cbytFebMonth + 1, 0)) = cbytFebLastDay Xor _ Day(DateSerial(Year(datDate2), cbytFebMonth + 1, 0)) = cbytFebLastDay Then ' Only one date falls within a leap year. ' Adjust both dates

This is expected because it is not possible to overflow values like "32 days" which could be interpreted as anything from "1 month and 4 days" to "1 month and 1 According to the human mind, this should be a difference of {June 25} which is 1 month, because July 25 has not occurred in the ending month, right? Here are several links that discuss computing age with the DATEDIFF function. Since age cannot be negative, we need another, more elaborate fix.

Keep it simple! That last point about data types is interesting; it isn't discussed in the Help, it wasn't mentioned in Documenting DATEDIFF, and it makes this Bug Hunt all the more worthwhile. I wish I knew VBA well enough to assist, but I don't. I was going to send this to you in a PM, but you do not have receive PMs from other members enabled.

OK means the DATEDIFF usage is OK given the data types involved. Wednesday, May 09, 2012 - 11:15:59 AM - Bill Back To Top Dave, I don't understand your question . . . Wednesday, May 09, 2012 - 11:51:12 AM - Joseph Back To Top Bill, I don't think the issue with the formula in question pertains to calculation of days, but specificallly when To start viewing messages, select the forum that you want to visit from the selection below.

AS int) instead */SELECT ABS(CAST((DATEDIFF(DAY,'2012-01-07','2012-01-06')/365.242199) AS int)) years;SELECT ABS(CAST((DATEDIFF(DAY,'2012-01-07','2012-01-06')/30.4375) AS int)) months;SELECT ABS(CAST(DATEDIFF(DAY,'2012-01-07','2012-01-06') AS int)) days; /* Multiple Year boundaries */SELECT FLOOR((DATEDIFF(DAY,'2008-01-07','2012-01-06')/365.242199)) years;SELECT FLOOR((DATEDIFF(DAY,'2008-01-07','2012-01-06')/30.4375)) months;SELECT FLOOR(DATEDIFF(DAY,'2008-01-07','2012-01-06')) days; /* Leap Year within With Day of Year, the number is inconsistent for dates after 2/28. The DATEDIFF function does not work that way. If I discover a fix, I will amend my solution and post it in a comment this evening.

The final expression even validly accounts for leap years. This function is available in all versions of Excel since version 5, but is documented only in the help files for Excel 2000. Also, the AgeInDays value is negative. You could avoid this by using Day of Non-Leap Year, meaning subtract 1 from leap years for dates over 59.

Wrapping the DATEDIFF function in the ABS allows the result to be positive whether the start date is before or after the end date. For example, if you only need to compute age as less than year or 1 year or more, you may be able to use a simpler solution than one that needs And the Year value of {} which is 0 full years. DECLARE @email_sent_at TIMESTAMP; DECLARE @current_timestamp TIMESTAMP DEFAULT CURRENT TIMESTAMP; ...