If day is missing need to represent the day of the month?

[rajeshalwayswel]

data have;
input date $ 10.;
cards;
2017-01-01
2016-10
2010-11
;
run;

Required:

2017-01-01
2016-10-31
2010-11-30

[novinosrin]

data have;
input date $ 10.;
cards;
2017-01-01
2016-10
2010-11
;
run;

data want;
set have;
new_date=input(date,anydtdte21.);
if length(date)=7 then new_Date=intnx('month',new_date,0,'e');
format new_date yymmdd10.;
run;

[Reeza · Accepted solution]

data want;
set have;

if length(date)<10 then date_num =intnx('month', input(cats(date, '-01'), yymmdd10.), 0, 'e');

else date_num = input(date, yymmdd10.);
format date_num date9.;


run;

[rajeshalwayswel]

Thanks to both. but the year is leap year in this scenario what would be the solution?

[novinosrin]

Please test and let me know

[rajeshalwayswel]

Thanks It working.

[Reeza]

Leap year doesn't matter.

---

@rajeshalwayswel wrote:

Thanks to both. but the year is leap year in this scenario what would be the solution?

---

[ballardw]

---

@rajeshalwayswel wrote:

Thanks to both. but the year is leap year in this scenario what would be the solution?

---

SAS knows which years are leap years, at least through 19999 the last time looked, and 'e' option in intnx takes that into account.

Yes I said 19999. I had to demonstrate this multiple times during Y2K worries about the data I was storing in SAS. The date formats don't display properly because of the 5-digit year but the date functions work just fine.