Date & Time Functions

Patrik Šimek Updated September 17, 2018 17:03

addDays (date; number)

Returns a new date as a result of adding a given number of days to a date. To subtract days, enter a negative number.

addDays(2016-12-08T15:55:57.536Z;2)
= 2016-12-10T15:55:57.536Z

addDays(2016-12-08T15:55:57.536Z;-2)
= 2016-12-6T15:55:57.536Z

addHours (date; number)

Returns a new date as a result of adding a given number of hours to a date. To subtract hours, enter a negative number.

addHours(2016-12-08T15:55:57.536Z;2)
= 2016-12-08T17:55:57.536Z

addHours(2016-12-08T15:55:57.536Z;-2)
= 2016-12-08T13:55:57.536Z

addMinutes (date; number)

Returns a new date as a result of adding a given number of minutes to a date. To subtract minutes, enter a negative number.

addMinutes(2016-12-08T15:55:57.536Z;2)
= 2016-12-08T15:57:57.536Z

addMinutes(2016-12-08T15:55:57.536Z;-2)
= 2016-12-08T15:53:57.536Z

addMonths (date; number)

Returns a new date as a result of adding a given number of months to a date. To subtract months, enter a negative number.

addMonths(2016-08-08T15:55:57.536Z;2)
= 2016-12-10T15:55:57.536Z

addMonths(2016-08-08T15:55:57.536Z;-2)
= 2016-12-06T15:55:57.536Z

addSeconds (date; number)

Returns a new date as a result of adding a given number of seconds to a date. To subtract seconds, enter a negative number.

addSeconds(2016-12-08T15:55:57.536Z;2)
= 2016-12-08T15:55:59.536Z

addSeconds(2016-12-08T15:55:57.536Z;-2)
= 2016-12-08T15:55:55.536Z

addYears (datum;years)

Returns a new date as a result of adding a given number of years to a date. To subtract years, enter a negative number.

addYears(2016-08-08T15:55:57.536Z;2)
= 2018-08-08T15:55:57.536Z

addYears(2016-08-08T15:55:57.536Z;-2)
= 2014-08-08T15:55:57.536Z

formatDate (date; format; [timezone])

Returns a date in the requested format and optionally, in a specified timezone. For example, format DD.MM.YYYY HH:mm. See the list of supported timezones.

formatDate(2016-12-28T16:03:06.372Z;MM/DD/YYYY)
= 12/28/2016

formatDate(2016-12-28T16:03:06.372Z;YYYY-MM-DD hh:mm A)
= 2016-12-28 4:03 PM

formatDate(2016-12-28T16:03:06.372Z;DD.MM.YYYY HH:mm;Europe/Prague)
= 28.12.2016 17:03

parseDate (date; format; [timezone])

Parses a string with a date and returns the date.

parseDate(2016-12-28;YYYY-MM-DD)
= 2016-12-28T00:00:00.000Z

parseDate(2016-12-28 16:03;YYYY-MM-DD HH:mm)
= 2016-12-28T16:03:00.000Z

parseDate(2016-12-28 04:03 pm;YYYY-MM-DD hh:mm a)
= 2016-12-28T16:03:06.000Z

parseDate(1482940986;X)
= 2016-12-28T16:03:06.000Z

Examples

How to calculate days between dates

One possibility is to employ the following expression:

​

You can copy & paste the following code:

{{round((2.value - 1.value) / 1000 / 60 / 60 / 24)}}

NOTE: Values of D1 and D2 have to be of type Date. If they are of type String (e.g. "20.10.2018"), use parseDate() function to convert them to type Date.

INFO: The round() function is used for cases when one of the dates falls within the daylight savings time period and the other not. In these cases, the difference in hours is by one hour less/more and dividing it by 24 gives a non-integer results.