DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0) Can someone explain me this

Question

DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0)

Can someone explain me this.

Answer 1

It's getting the start of the current month. It counts the number of months since January 1900, and adds on them to January 1 1900 to get the start of the current month

Answer 2

The DATEDIFF will give you date diff in month from January 1 1900 to current date
AND 
The DATEADD will add (DATEDIFF) results months to your last parameter of DATEADD

Answer 3

The DateDiff function returns how many seconds, months, years - whatever interval you specify between the first date (here 0) and the second date (here the current date).

     DATEDIFF(MONTH, 0, '2-14-2015')  --returns month. 0 is for 1/1/1900, and getdate is the current date 
     --(i used a set date bc dates will change as this post gets older).
     result: 1381

In my workings, I used the date of 2-14-2015 and it gave me 1381 number of months since 1/1/1900. I put 1381 into the dateadd function like so...

     select Dateadd(MONTH, 1381, 0)
     result: 2015-02-01 00:00:00.000

Basically, it gets the first day of the whatever month is specified in the date of the innermost formula.

The code i was having to figure out went a step farther and subtracted 1 second from that result to get the last day of the month at 11:59:59 like so...

     select DATEADD(s, -1, '2015-02-01 00:00:00.000')

The Formula all put together looks like this:

     DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,getdate())+1,0))

Hope this helps someone. :)

Answer 4

this will give you the first of the month for a given date

inner select select DATEDIFF(MONTH, 0, GETDATE()) will give the number of months from 1900-01-01

here it is 1350

this will be add to 1900-01-01 , but only the months

select DATEADD(MONTH,1350,0) will give 2012-07-01 00:00:00.000

which is the start of the current month.

I think this is the most efficient way to find the starting of a month for any given date.

Answer 5

The DATEADD function adds an interval to a date you specify. For example, if the due dates of all orders in the SalesOrderHeader table slipped 3 days, you could obtain the new dates with the following statement:

USE AdventureWorks;
GO
SELECT DATEADD(day, 3, DueDate)
FROM Sales.SalesOrderHeader;
GO

The DATEDIFF function calculates the period of time in dateparts between the second and first of two dates you specify. In other words, it finds an interval between two dates. The result is a signed integer value equal to date2 - date1 in date parts. The following query uses the date November 30, 2001, and finds the number of days that elapsed between DueDate and that date:

USE AdventureWorks;
GO
SELECT DATEDIFF(day, DueDate, 'Nov 30 2001')
FROM Sales.SalesOrderHeader;
GO



SELECT DATEDIFF(year, '20051220', '20060101')
SELECT DATEDIFF(month, '20051220', '20060101')
SELECT DATEDIFF(day, '20051220', '20060101')