Date & Time Functions

Patrik Šimek Updated June 17, 2020 11:04

formatDate (date; format; [timezone])

When to use it: You have a Date value that you wish to convert (format) to a Text value (textual human readable representation) like 12-10-2019 20:30 or Aug 18, 2019 10:00 AM

Parameters

The second column indicates the expected type. If different type is provided, Type Coercion is applied.

Return value and type

Text representation of the given Date value according to the specified format and timezone. Type is Text.

Examples

The Organization's and Web's timezone were both set to Europe/Prague in the following examples.

formatDate(1. Date created;MM/DD/YYYY)
= 10/01/2018

formatDate(1. Date created;YYYY-MM-DD hh:mm A)
= 2018-10-01 09:32 AM

formatDate(1. Date created;DD.MM.YYYY HH:mm;UTC)
= 01.10.2018 07:32

formatDate(now;DD.MM.YYYY HH:mm)
= 19.03.2019 15:30

parseDate (text; format; [timezone])

When to use it: You have a Text value representing a date (e.g. 12-10-2019 20:30 or Aug 18, 2019 10:00 AM) and you wish to convert (parse) it to a Date value (binary machine readable representation).

Parameters

The second column indicates the expected type. If different type is provided, Type Coercion is applied.

Return value and type

Date representation of the given Text value according to the specified format and timezone. Type is Date.

Examples

Please note that in the following examples the returned Date value is expressed according to ISO 8601, but the actual resulting value is of type 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

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)
= 2016-10-08T15:55:57.536Z

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

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

setSecond (date; number)

Returns a new date with the seconds specified in parameters. Accepts numbers from 0 to 59. If a number is given outside of this range, it will return the date with the seconds from the previous or subsequent minute(s), accordingly.

setSecond(2015-10-07T11:36:39.138Z;10)
= 2015-10-07T11:36:10.138Z

setSecond(2015-10-07T11:36:39.138Z;61)
= 2015-10-07T11:37:01.138Z

setMinute (date; number)

Returns a new date with the minutes specified in parameters. Accepts numbers from 0 to 59. If a number is given outside of the range, it will return the date with the minutes from the previous or subsequent hour(s), accordingly.

setMinute(2015-10-07T11:36:39.138Z;10)
= 2015-10-07T11:10:39.138Z

setMinute(2015-10-07T11:36:39.138Z;61)
= 2015-10-07T12:01:39.138Z

setHour (date; number)

Returns a new date with the hour specified in parameters. Accepts numbers from 0 to 23. If a number is given outside of the range, it will return the date with the hour from the previous or subsequent day(s), accordingly.

setHour(2015-08-07T11:36:39.138Z;6)
= 2015-08-07T06:36:39.138Z

setHour(2015-08-07T11:36:39.138Z;-6)
= 2015-08-06T18:36:39.138Z

setDay (date; number/name of the day in english)

Returns a new date with the day specified in parameters. It can be used to set the day of the week, with Sunday as 1 and Saturday as 7. If the given value is from 1 to 7, the resulting date will be within the current (Sunday-to-Saturday) week. If a number is given outside of the range, it will return the day from the previous or subsequent week(s), accordingly.

setDay(2018-06-27T11:36:39.138Z;monday)
= 2018-06-25T11:36:39.138Z

setDay(2018-06-27T11:36:39.138Z;1)
= 2018-06-24T11:36:39.138Z

setDay(2018-06-27T11:36:39.138Z;7)
= 2018-06-30T11:36:39.138Z

setDate (date; number)

Returns a new date with the day of the month specified in parameters. Accepts numbers from 1 to 31. If a number is given outside of the range, it will return the day from the previous or subsequent month(s), accordingly.

setDate(2015-08-07T11:36:39.138Z;5)
= 2015-08-05T11:36:39.138Z

setDate(2015-08-07T11:36:39.138Z;32)
= 2015-09-01T11:36:39.138Z

setMonth (date; number/name of the month in English)

Returns a new date with the month specified in parameters. Accepts numbers from 1 to 12. If a number is given outside of this range, it will return the month in the previous or subsequent year(s), accordingly.

setMonth(2015-08-07T11:36:39.138Z;5)
= 2015-05-07T11:36:39.138Z

setMonth(2015-08-07T11:36:39.138Z;17)
= 2016-05-07T11:36:39.138Z

setMonth(2015-08-07T11:36:39.138Z;january)
= 2015-01-07T12:36:39.138Z

setYear (date; number)

Returns a new date with the year specified in parameters.

setYear(2015-08-07T11:36:39.138Z;2017)
= 2017-08-07T11:36:39.138Z

Examples

How to calculate n-th day of week in month

[adapted for Integromat from this original source]

If you need to calculate a date corresponding to n-th day of week in month (e.g. 1st Tuesday, 3rd Friday, etc.), you may use the following formula:

You may copy and paste the formula's code into a field:

{{addDays(setDate(1.date; 1); 1.n * 7 - formatDate(addDays(setDate(1.date; 1); "-" + 1.dow); "E"))}}

The formula contains the following items:

In case you wish to calculate only one specific case, e.g. 2nd Wednesday, you may replace the items 1.n and 2.dow in the formula with corresponding numbers. For 2nd Wednesday in current month you would use the following values:

• 1.n = 2
• 1.dow = 3
• 1.date = now

Explanation:

• setDate(now;1) returns first of current month
• formatDate(....;E) returns day of week (1, 2, ... 6)
• see the original source for the rest

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.

How to calculate last day/millisecond of month

When specifying a date range (e.g. in a search module) spanning the whole previous month as closed interval (the interval that includes both its limit points), it is necessary to calculate last day of month.

2019-09-01 ≤ D ≤ 2019-09-30

The formula below shows one way how to calculate last day of the previous month:

You can copy & paste the following code:

{{addDays(setDate(now; 1); -1)}}

In some cases, it is necessary to calculate not only the last day of month, but literally its last millisecond:

2019-09-01T00:00:00.000Z ≤ D ≤ 2019-09-30T23:59:59.999Z

The formula below shows one way how to calculate last millisecond of the previous month:

You can copy & paste the following code:

{{parseDate(parseDate(formatDate(now; "YYYYMM01"); "YYYYMMDD"; "UTC") - 1; "x")}}

If the result should respect your timezone settings, simply omit the UTC argument:

{{parseDate(parseDate(formatDate(now; "YYYYMM01"); "YYYYMMDD") - 1; "x")}}

However, it is preferable to use half-open interval instead (the interval that excludes one of its limit points), specifying the first day of the following month instead and replacing the less or equal than operator with less than:

2019-09-01 ≤ D < 2019-10-01

2019-09-01T00:00:00.000Z ≤ D < 2019-10-01T00:00:00.000Z