Character to Date

[ZZB]

How to convert the following Charater to Date format?

For example, raw variable DATE "01 MAR 2013 00:00:02.358". The length is 24. There are space between '01' and 'MAR' and '2013' and '00'.

How to seperate it into two column '03/01/2013' + 'hh:mm:ss:sss' or 'hh:mm:ss'. Both are date format. Thanks!

[Reeza]

Are you reading these from a text file or is the data already in SAS?

You can try compress and input to read in the data.

untested:

datetimevar=input(compress(char_val), anydtdtm.);
var_datepart=datepart(datetimevar);
var_timepart=timepart(datetimevar);

format var_datepart date9. var_timepart time8.;

[ZZB]

The data is already in SAS.
datetimevar=input(compress(char_val), anydtdtm.); DOES NOT WORK.
samnan's code works. Thanks!

[Reeza]

The informat provided by @data_null__ is the correct informat. Correcting the informat fixes the issues.

data have;

original_var = "01 MAR 2013 00:00:02.358";
temp=compress(original_var);

dt_var=input(original_var, datetime24.3);
date_part=datepart(dt_var);
time_part=timepart(dt_var);

format dt_var datetime24.3 date_part date9. time_part time11.3;
run;

proc print data=have;
run;

[data_null__]

34         data _null_;
35            input dt & datetime.;
36            put (dt)(=datetime22.3);
37            list;
38            cards;

dt=01MAR2013:00:00:02.358
RULE:      ----+----1----+----2----+----3----+----4----+----5----+----6----+----7----+----8----+----9----+----0                     
39         01 MAR 2013 00:00:02.358

[samnan · Accepted solution]

Hi ZZB;

try this 

data have;
DATE = "01 MAR 2013 00:00:02.358";

run;

data want;
set have;
NewDate = input(compress (substr(DATE ,1,11)),date9.);
NewTime = input( scan(DATE,-1,' '),time12.);
format NewDate date9. NewTime time12.;

run;