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  1. datepart. Un literal o cadena de identificación de la parte específica del valor de la fecha (por ejemplo, año, mes o día) en la que actúa la función. Para obtener más información, consulte Partes de fecha para funciones de fecha o marca temporal. {date|timestamp}

  2. Syntax. DATE_PART( datepart, {date | timestamp }) Arguments. datepart. An identifier literal or string of the specific part of the date value (for example, year, month, or day) that the function operates on. For more information, see Date parts for date or timestamp functions. {date|timestamp}

  3. Guía para desarrolladores de bases de datos. Función DATE_PART_YEAR. PDF RSS. La función DATE_PART_YEAR extrae el año de una fecha. Sintaxis. DATE_PART_YEAR( date) Argumento. date. Una columna de tipo de datos DATE o una expresión que toma implícitamente el valor de un tipo DATE. Tipo de retorno. INTEGER. Ejemplos.

  4. Explore Redshift's date functions: now(), sysdate(), current_date, and current_time. Learn how to perform date and time manipulations, find rows within specific date ranges, extract timestamp components, convert timestamps to Unix time, and calculate time differences using functions.

  5. 7 de dic. de 2022 · Date part is an argument used in Redshift date functions. It is a single lowercase word ( datepart) used to denote a part of a date. This could be day, month, year and so on. The value for the datepart argument is specified without quotes and in lowercase. For example, month is specified as mon.

  6. 18 de may. de 2022 · SELECT DATEPART(week, date(date)), MIN(date) in Redshift with the start day being Saturday. I tried using different syntax for SET DATEFIRST but it seems like it's not supported (or I just can't find the correct syntax for redshift).

  7. Hace 6 días · Redshift Date functions: DATE_DIFF function. This function takes two date values and returns the difference between them in the unit specified by the ‘date part’ argument. For example, to find the difference between two dates in weeks, use the below statement. SELECT DATEDIFF(week,'2019-01-01','2019-12-31') as numweeks;