DATEDIF Function

Calculates the difference between two valid date values for the specified units of measure.

  • Inputs must be column references.
  • The first value is used as the baseline to compare the date values.
  • Results are calculated to the integer value that is closest to and lower than the exact total; remaining decimal values are dropped.

Basic Usage

derive value:DATEDIF(StartDate, EndDate, month)

Output: Generates a column of values calculating the number of full months that have elapsed between StartDate and EndDate.


derive value:DATEDIF(date1,date2,date_units )

ArgumentRequired?Data TypeDescription
date1YdatetimeStarting date to compare
date2YdatetimeEnding date to compare
date_unitsYstringString literal representing the date units to use in the comparison

For more information on syntax standards, see Language Documentation Syntax Notes.

date1, date2

Date values to compare using the date_units units. If date2 > date1, then results are positive.

  • Date values must be column references.
  • If date1 and date2 have a specified time zone offset, DATEDIF calculates the difference including the timezone offsets.
  • If date1 does not have a specified time zone but date2 does, DATEDIF uses the local time in the same time zone as date2 to calculate the difference. DATEDIF returns the difference without the time zone offset.

Usage Notes:

Required?Data TypeExample Value
YesString (Date column reference)LastContactDate


Unit of date measurement to calculate between the two valid dates.

Usage Notes:

Required?Data TypeExample Value

Accepted Value for date units:

  • year
  • quarter
  • month
  • dayofyear

  • week
  • day
  • hour
  • minute
  • second
  • millisecond


Example - aged orders

This example illustrates how to use the DATEDIF function to calculate the number of days that have elapsed between the order date and today for purposes of informing the customer.


For the orders in the following set, you want to charge interest for those ones that are older than 90 days.



The first step is to create a column containing today's (3/16/16) date value:

derive value:TODAY() as:'Today'

You can now use this value as the basis for computing the number of elapsed days for each invoice:

derive value:DATEDIF(OrderDate, Today, day)

The age of each invoice in days is displayed in the new column. Now, you want to add a little bit of information to this comparison. Instead of just calculating the number of days, you could write out the action to undertake. Replace the above with the following:

derive value:IF((DATEDIF(OrderDate, Today, day) > 90),'Charge interest','no action') as:'TakeAction'

To be fair to your customers, you might want to issue a notice at 45 days that the invoice is outstanding. You can replace the above with the following:

derive value:IF(DATEDIF(OrderDate, Today, day) > 90,'Charge interest',IF(DATEDIF(OrderDate, Today, day) > 45),'Send letter','no action')) as: 'TakeAction'

By using nested instances of the IF function, you can generate multiple results in the TakeAction column.

For the items that are over 90 days old, you want to charge 5% interest. You can do the following:

set col:Amount value:IF(TakeAction == 'Charge interest',Amount * 1.05,Amount)

The above sets the value in the Amount column based on the conditional of whether the TakeAction column value is Charge interest. If so, apply 5% interest to the value in the Amount column.


10011/31/16100003/03/16no action
100211/15/15105003/03/16Charge interest
100312/18/15100003/03/16Send letter
10041/15/16100003/03/16Send letter

Example - dayofyear Calculations

This example demonstrates how dayofyear is calculated using the DATEDIF function, specifically how leap years and leap days are handled. Below, you can see some example dates. The year 2012 was a leap year.


11/1/1010/10/10Same year; no leap year
21/1/1010/10/11Different years; no leap year
310/10/111/1/10Reverse dates of previous row
42/28/114/1/11Same year; no leap year;
52/28/124/1/12Same year; leap year; spans leap day
62/29/124/1/12Same year; leap year; d1 = leap day
72/28/112/29/12Diff years; d2 = leap day; converted to March 1 in d1 year


In this case, the transform is simple:

derive value:DATEDIF(d1,d2,dayofyear) as:'datedifs'


11/1/1010/10/10282Same year; no leap year
21/1/1010/10/11282Different years; no leap year
310/10/111/1/10-282Reverse dates of previous row
42/28/114/1/1132Same year; no leap year;
52/28/124/1/1233Same year; leap year; spans leap day
62/29/124/1/1232Same year; leap year; d1 = leap day
72/28/112/29/121Diff years; d2 = leap day; converted to March 1 in d1 year

Rows 1 - 3:

  • Row 1 provides the baseline calc.
  • In Row 2, the same days of the year are used, but the year is different by a count of 1. However, since we are computing dayofyear the result is the same as for Row 1.

    NOTE: When computing dayofyear, the year value for d2 is converted to the year of d1. The difference is then computed.

  • Row 3 represents the reversal of dates in Row 2.

    NOTE: Negative values for a dayofyear calculation indicate that d2 occurs earlier in the calendar than d1, ignoring year.

Rows 4 - 7: Leap years

  • Row 4 provides a baseline calculation for a non-leap year.
  • Row 5 uses the same days of year as Row 4, but the year (2012) is a leap year. Dates span a leap date (February 29). Note that the DATEDIF result is 1 more than the value in the previous row.

    NOTE: When the two dates span a leap date and the year for d1 is a leap year, then February 29 is included as part of the calculated result.

  • Row 6 moves date 1 forward by one day, so it is now on a leap day date. Result is one less than the previous row, which also spanned leap date.
  • Row 7 switches the leap date to d2. In this case, d2 is converted to the year of d1. However, since it was a leap day originally, in the year of d1, this value is March 1. Thus, the difference between the two dates is 1.

    NOTE: If d2 is a leap date and the year for d1 is not a leap year, the date used in for d2 is March 1 in the year of d1.

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