date extract

[Eugenio211]

Hi,

how to extract date from this type of date

' 2018-08-26 17:51:30.8770000 +00:00'?

Thank you.

[PaigeMiller]

If this is a numeric value in a numeric variable, what you are showing (without the quotes) is a date/time value, and so the DATEPART function extracts the date.

If it is character ... well ... it shouldn't be. Date values and date/time values should always be numeric.

What does PROC CONTENTS show for this variable?

[Eugenio211]

this is what I get when using datepart:
ERROR: Function DATEPART requires a numeric expression as argument 1.
ERROR: Character expression requires a character format.

[Eugenio211]

this is my step:
proc sql;
create table PaidAhead as
select l.CUSTOMER_KEY,l.LOAN_KEY,l.DATE_ENTERED
,datepart(PAYOFF_DATE) as PAID_OFF_DATE format date9.
,l.product_type
from DataSet1 L
join DataSet2 d on l.LOAN_KEY = d.LOAN_KEY
;quit;

[PaigeMiller]

Obvi you have a character variable that looks like a date/time; but it clearly is not a date/time since date/time values cannot be character.

You have to convert this to numeric, with INPUT function and the proper informat.

data want;
    datetime=' 2018-08-26 17:51:30.8770000 +00:00';
    datetime_numeric=input(datetime, anydtdtm.);
    date=datepart(datetime_numeric);
    format date date9.;
run;

Please do yourself a favor and do not use character strings for dates, times and datetimes. These variables should be NUMERIC, always, 100% of the time, no exceptions.

[Astounding]

Perhaps a little simpler:

data want;
   set have;
   date = input(big_long_datetime_string, : yymmdd10.);
   drop big_long_datetime_string;
   formate date yymmdd10.;
run;

This gets you the DATE as a proper, numeric SAS date, while getting rid of the time part.  If memory serves, however, you will need to use a DATA step because SQL will not allow the INPUT function.  (Unless times have changed, but I doubt it.)