Topic: Datediff Function
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How to use Datediff function
Dim date1 As DateTime = Now Dim date2 As DateTime = "01/01/1900" Dim numMonths As Integer = DateDiff(DateInterval.Month, date2, date1) Dim numYears As Integer = DateDiff(DateInterval.Year, date2, date1) Dim n... Read More »
Source: http://www.experts-exchange.com/Q_23166300.htm
How to use DateDiff function in Java?
Date date1;// = // get your first date Date date2;// = // get your second date Calendar c = Calendar.getInstance(); Calendar c2 = Calendar.getInstance(); c... Read More »
Source: http://www.experts-exchange.com/Programming/Languages/Java/Q_2027...
What is wrong with this DateDiff function in an Access VB query
PAQed with points refunded (500) DarthMod Community Support Moderator Read More »
Source: http://www.experts-exchange.com/Q_21876937.htm
More Common Questions
Answers to Other Common Questions
That looks right, but it's obviously giving you weeks instead of weekdays. Aha, it's a known bug: http://support.microsoft.com/kb/207795
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Source: http://www.experts-exchange.com/Q_24786866.html
In SQL Server 2005, do not encapsulate the hh in single quotes UPDATE ActiveUsers SET LogOut = { fn NOW() }, TimeLogged = DATEDIFF(hh, DatetimeLogged, GETDATE()) WHERE (UserID = 2)
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Source: http://www.experts-exchange.com/Q_23466115.htm
If not, here is a custom tag that should <!--- BusDateDiff.cfm Description: This CF custom tag calculates the number of business days between two dates. Attributes: START - (required) Start date. END - (optional) End Date. Default is Now()....
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Source: http://www.experts-exchange.com/Software/Server_Software/Web_Serv...
The DateDiff function returns the number of intervals between two dates. DateDiff (interval, date1, date2 [,firstdayofweek [,firstweekofyear]]) Interval (Required) - The interval you want to use to calculate the differences between date1 an...
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Source: http://answers.yahoo.com/question/index?qid=20080222130909AANPrMY
don't know where in your code you want the leap year, assuming you want to check your date1: year=date1.getYear()+1900; if ((year%4==0)^(year%100==0)^(year%400==0)) { alert(year+' is a leap year'); } > but what is ^ exclisive or > .. i know...
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Source: http://www.experts-exchange.com/Programming/Languages/Scripting/J...
select convert(varchar(20), getdate(), 101)
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Source: http://www.experts-exchange.com/Software/Office_Productivity/Offi...
Maybe you'll like the rxlib ? get it at www.rxlib.com, freeware with source. Look at the dateUtils unit : function CurrentYear: Word; function IsLeapYear(AYear: Integer): Boolean; function DaysPerMonth(AYear, AMonth: Integer): Integer; func...
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Source: http://www.experts-exchange.com/Programming/Languages/RAD/Delphi/...
