adding days to start date

[monday89]

I have the following dataset. What I would like to do is that if day is blank then add 7 days to start_dtm

ptid	days	start_dtm
1		03DEC19:04:35:00

Something like this:

ptid	days	start_dtm	stop_dtm
1		03DEC19:04:35:00	10DEC19:04:35:00

I tried both codes and doesn't work:

if missing(days) then stop_dtm = intnx('dtday', start_dtm, 7);

ALSO

if missing(days) then stop_dtm =  start_dtm + 7*24*60*60

but i am getting 01Jan20:23:59 for stop date

[ballardw]

I suspect that you want:

if missing(days) then stop_dtm = intnx('dtday', start_dtm, 7,'S');

the S, alignment, in the last position means "same", as in the same hour and minute. Default is 'B' or begin for the Intnx function would would advance to the start of the day or 00:00:00

(you didn't show the result you were getting).

Hint: you may want to make sure that your dates display with 4 digit years.

And without your actual values and the complete code of a data step it really isn't possible to tell why something goes "wrong".

Doesn't work is awful vague.

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[mklangley · Accepted solution]

Try running the code I included. It looks like both those approaches will add seven days. (The second one will ensure the timestamp is correct.)

How are you getting 01Jan20:23:59 for the stop date? Could you include the exact code you ran for that?

data have;
    input ptid days start_dtm :datetime16.;
    format start_dtm datetime16.;
    datalines;
    1 . 03DEC19:04:35:00
    ;
run;

data want;
    set have;
    format stop_dtm stop_dtm2 datetime16.;
    if missing(days) then stop_dtm = intnx('dtday', start_dtm, 7);
    if missing(days) then stop_dtm2 =  start_dtm + 7*24*60*60;
run;