Example
Return a specified part of a date:
sql
SELECT DATEPART(year, '2017/08/25') AS DatePartInt;
Definition and Usage
The DATEPART() function returns a specified part of a date.
This function returns the result as an integer value.
Syntax
text
DATEPART(interval, date)
Parameter Values
| Parameter | Description |
|---|---|
| interval | Required. The part of the date argument to return. Can be one of the following values: year, yyyy, yy = Year quarter, qq, q = Quarter month, mm, m = month dayofyear, dy, y = Day of the year day, dd, d = Day of the month week, ww, wk = Week weekday, dw, w = Weekday hour, hh = hour minute, mi, n = Minute second, ss, s = Second millisecond, ms = Millisecond microsecond, mcs = Microsecond nanosecond, ns = Nanosecond tzoffset, tz = Timezone offset iso_week, isowk, isoww = ISO week |
| date | Required. The date to use. Can be one of the following data types: date, datetime, datetimeoffset, datetime2, smalldatetime, or time |
Technical Details
| Return type: | int |
|---|---|
| Works in: | SQL Server (starting with 2008), Azure SQL Database, Azure SQL Data Warehouse, Parallel Data Warehouse |
More Examples
Example
Return a specified part of a date:
sql
SELECT DATEPART(yy, '2017/08/25') AS DatePartInt;
Example
Return a specified part of a date:
sql
SELECT DATEPART(month, '2017/08/25') AS DatePartInt;
Example
Return a specified part of a date:
sql
SELECT DATEPART(hour, '2017/08/25 08:36') AS DatePartInt;
Example
Return a specified part of a date:
sql
SELECT DATEPART(minute, '2017/08/25 08:36') AS DatePartInt;