topic problems when converting datetime character to date9. if missing seconds in SAS Programming

problems when converting datetime character to date9. if missing seconds

noda6003 — Mon, 21 Jan 2019 10:13:31 GMT

I have a datetime character variable which i am converting to date9. but for few obs seconds are missing

ldt atd

2018-12-13T10:30:00 13DEC2018

2018-12-10T09:40 .

I use the this code but it makes the second observation mentioned above as missing

input(ldt, anydtdtm.) as atd format=dtdate9.

is there any way i can make ldt =2018-12-10T09:40 as 2018-12-10T09:40:00

Please suggest

Re: problems when converting datetime character to date9. if missing seconds

PeterClemmensen — Mon, 21 Jan 2019 10:21:52 GMT

Just to be clear, do you want Adt to be a datetime variable or a date variable?

Re: problems when converting datetime character to date9. if missing seconds

PeterClemmensen — Mon, 21 Jan 2019 10:24:11 GMT

If you want it to be a date variable, simply extract the date part of the character variable and convert like this

data have;
ldt="2018-12-13T10:30:00";output;
ldt="2018-12-10T09:40";output;
run;

data want;
   set have;
   Adt=input(substr(ldt, 1, 10), yymmdd10.);
   format Adt date9.;
run;

Re: problems when converting datetime character to date9. if missing seconds

novinosrin — Mon, 21 Jan 2019 14:17:19 GMT

No need to extract the chars as th very informat's instructions yymmdd10. will read the non standard char date upto 10 chars and convert to a sas date.

So just apply the informat-->

data have;
ldt="2018-12-13T10:30:00";output;
ldt="2018-12-10T09:40";output;
run;

data want;
   set have;
   Adt=input(ldt, yymmdd10.);
   format Adt date9.;
run;