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DATEPART() Function in SQL Server

  • Last Updated : 18 Jan, 2021

DATEPART() function :
This function in SQL Server is used to find a given part of the specified date. Moreover, it returns the output value as an integer.

Features :

  • This function is used to find a given part of the specified date.
  • This function comes under Date Functions.
  • This function accepts two parameters namely interval and date.
  • This function can also include time in the date section.
  • This function returns the output in integer form.

Syntax :

DATEPART(interval, date)

Parameter :
This method accepts two parameters as given below as follows.

  • interval – 
    It is the specified part which is to be returned. Moreover, the values of the interval can be as given below.
year, yyyy, yy      = Year, which is the specified year.
quarter, qq, q      = Quarter, which is the specified quarter.
month, mm, m        = month, which is the specified month.
dayofyear, dy, y    = Day of the year, which is the specified day of the year.
day, dd, d          = Day, which is the specified day.
week, ww, wk        = Week, which is the specified week.
weekday, dw, w      = Weekday, which is the specified week day.
hour, hh            = hour, which is the specified hour.
minute, mi, n       = Minute, which is the specified minute.
second, ss, s       = Second, which is the specified second.
millisecond, ms     = Millisecond, which is the specified millisecond.
  • date – 
    It is the specified date which is to be used.

Returns :
It returns a given part of the specified date.

Example-1 :

Using DATEPART() function and getting the year part of the specified date.

SELECT DATEPART(year, '2017/08/25');

Output :

2017

Example-2 :
Using DATEPART() function and getting the month part of the specified date.

SELECT DATEPART(month, '2017/08/25');

Output :

8

Example-3 :
Using DATEPART() function and getting the day part of the specified date.

SELECT DATEPART(day, '2017/08/25');

Output :

25

Example-4 :
Using DATEPART() function and getting the hour part of the specified date which includes time as well.

SELECT DATEPART(hour, '2021/01/06 05:30');

Output :

5

Example-5 :
Using DATEPART() function and getting the second part of the specified date which includes time as well using a variable.

DECLARE @date VARCHAR(50);
SET @date = '2019/06/05 07:37:54';
SELECT DATEPART(second, @date);

Output :

54

Application :
This function is used to find the given part of the specified date.
 

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