DatePart()

Extracts a date part (such as the day, month, or year) from a date or timestamp string.

1DatePart(dateString, datePart)

The DatePart() function has two parameters:

• dateString (string): Required. A string that contains a date or timestamp.
• datePart (string): Required. The date part that you want to extract from the string. Accepted values: Y (years), M (months), D (days), H (hours), and MI (minutes).

You can specify a date string in several formats, including:

• ISO 8601 timestamp: 2023-08-05T13:41:23-06:00
• ISO 8601 date: 2023-08-05
• US date and time notation: 8/5/2023 1:41 PM
• Long-form notation: 5 August 2023 or August 5, 2023
• Date and time: 2023-08-05 1:41:23 PM
• Time only: 1:41 PM
• Chinese and Japanese notation: 2023 年 8 月 5 日
• Korean notation: 2023년 8월 5일

The function doesn’t support these formats:

• Dates in which the day includes an ordinal suffix, such as August 5th, 2023 or 5th August 2023.
• Little-endian numeral notation, such as 5/8/2023 to represent the 5th of August, 2023.
• Month names in languages other than English, such as 5 août 2023.
• Dates that include any numerals other than Western Arabic numerals, such as ٢٠٢٣/٨/٥.
• Dates based on any calendar system other than the Gregorian calendar, such as 18 Av, 5783 or 18 Muharram, 1445.

This example extracts the year from a date string.

1%%[
2  Var @date, @year
3  Set @date = "2023 年 8 月 5 日"
4  Set @year = DatePart(@date, "Y")
5]%%
6<p>The current year is %%=v(@year)=%%.</p>

The example outputs the current year:

1The current year is 2023.