Not applicable
Posts: 0

# Considering leap days.

My code:

data dates;
start='20040228'
tmpstart=input(start,yymmdd8.)+364;
end=put(tmpstart,yymmddn8.);
run;

Result: end='20050226'. That's what I want, since the start was the next to last day in february.

start='20040229' gives end='20050227'. OK!
start='20040301' gives end='20050228'. OK!

But: start='20030301' gives end ='20040228'. I want end='20040229'.

Is there some time function that gives start + one calendar year - 1 day?
Regular Contributor
Posts: 151

## Re: Considering leap days.

Hi
Use INTNX:

[pre]
data _null_;
start='20030301';
end=put(intnx('year',input(start,yymmdd8.),+1,'sameday')-1,yymmddn8.);
put start= end=;
run;
[/pre]
Frequent Contributor
Posts: 139

## Re: Considering leap days.

I would suggest one of the Date Time Functions in SAS such as INTNX.

http://support.sas.com/documentation/cdl/en/lrdict/63026/HTML/default/viewer.htm#a000245860.htm

Below is an example;

data dates;
length start \$8;
input start \$;
datalines;
20040228
20040229
20040301
20030301
;
run;

data one;
set dates;

tmpstart=input(start,yymmdd8.)+364;
end=put(tmpstart,yymmddn8.);

new_end=intnx('year',input(start,yymmdd8.),1,'sameday' )-1;
format new_end yymmdd10.;

put _all_;
run;
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