D toc 

Excerpt 

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.
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derive type:single value:DATEDIF(StartDate, EndDate, month) 
Output: Generates a column of values calculating the number of full months that have elapsed between StartDate
and EndDate
.
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derive type:single value:DATEDIF(date1,date2,date_units ) 
Argument  Required?  Data Type  Description 

date1  Y  datetime  Starting date to compare 
date2  Y  datetime  Ending date to compare 
date_units  Y  string  String literal representing the date units to use in the comparison 
D s lang 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
anddate2
have a specified time zone offset,DATEDIF
calculates the difference including the timezone offsets.  If
date1
does not have a specified time zone butdate2
does,DATEDIF
uses the local time in the same time zone asdate2
to calculate the difference.DATEDIF
returns the difference without the time zone offset.
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Required?  Data Type  Example Value 

Yes  String (Date column reference)  LastContactDate 
date_units
Unit of date measurement to calculate between the two valid dates.
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Required?  Data Type  Example Value 

Yes  String  year 
Accepted Value for date units:
year
quarter
 month
dayofyear
week
day
hour
minute
second
millisecond
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Example  aged orders
Include Page  


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.
Source:
dateId  d1  d2  Notes 

1  1/1/10  10/10/10  Same year; no leap year 
2  1/1/10  10/10/11  Different years; no leap year 
3  10/10/11  1/1/10  Reverse dates of previous row 
4  2/28/11  4/1/11  Same year; no leap year; 
5  2/28/12  4/1/12  Same year; leap year; spans leap day 
6  2/29/12  4/1/12  Same year; leap year; d1 = leap day 
7  2/28/11  2/29/12  Diff years; d2 = leap day; converted to March 1 in d1 year 
Transform:
In this case, the transform is simple:
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derive type:single value:DATEDIF(d1,d2,dayofyear) as:'datedifs' 
dateId  d1  d2  datedifs  Notes 

1  1/1/10  10/10/10  282  Same year; no leap year 
2  1/1/10  10/10/11  282  Different years; no leap year 
3  10/10/11  1/1/10  282  Reverse dates of previous row 
4  2/28/11  4/1/11  32  Same year; no leap year; 
5  2/28/12  4/1/12  33  Same year; leap year; spans leap day 
6  2/29/12  4/1/12  32  Same year; leap year; d1 = leap day 
7  2/28/11  2/29/12  1  Diff 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.Info NOTE: When computing
dayofyear
, the year value ford2
is converted to the year ofd1
. The difference is then computed.Row 3 represents the reversal of dates in Row 2.
Info NOTE: Negative values for a
dayofyear
calculation indicate thatd2
occurs earlier in the calendar thand1
, ignoring year.
Rows 4  7: Leap years
 Row 4 provides a baseline calculation for a nonleap 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.Info 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 ofd1
. However, since it was a leap day originally, in the year ofd1
, this value is March 1. Thus, the difference between the two dates is1
.Info NOTE: If
d2
is a leap date and the year ford1
is not a leap year, the date used in ford2
is March 1 in the year ofd1
.
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