The following code for outputing the date in uts format gives the data of Jan 1, 1960 instead of current date.
If keep the format as say date9. I get the correct value.
DATA createdates;
current1= date();
format current1 b8601dt.;
RUN;
PROC PRINT data=createdates;
title 'The createdates data set';
var current1 ;
RUN;:
The output is
The createdates data set 13:25 Wednesday, June 7, 2017 1
Obs current1
1 19600101T054937
Can anybody help?
change it to
DATA createdates; current1= datetime(); format current1 b8601dt.; RUN;
Looks like you're applying a datetime format to a date value.
change it to
DATA createdates; current1= datetime(); format current1 b8601dt.; RUN;
Thanks
it worked
The createdates data set 13:58 Wednesday, June 7, 2017 1
Obs current1
1 20170607T135842
I there a way can remove the T and have the values like "20170607135842".
The idea is to have it in YYYYMMDDHHmmSS?
you may to have to use proc format with picture statement as shown below
PROC FORMAT; picture dt other='%0Y-%0m-%0d%0H%0M%0S' (datatype=datetime); RUN; DATA createdates; current1= put(datetime(),dt.); RUN;
.
Thanks Kiran.
I appreciate your time and efforts in replying to me and reminding that datetime is the proper format to use.
My objective was to get date in the format YYYYMMDDHHmmSS.
So the last solution is not what I want.
I had to remove T from the output. So I used this. compress(put(datetime(),b8601dt.),'T');
For comparison I ran your and my solution
PROC FORMAT;
picture dt other='%0Y-%0m-%0d%0H%0M%0S' (datatype=datetime);
RUN;
DATA createdates;
current1= put(datetime(), dt.);
todays_date=compress(put(datetime(),b8601dt.),'T');
RUN;
and got the result as follows
current1 = 2017-06-07215453
todays_date= 20170607215454
There could be many other ways to do ,
I am accepting your earlier post as a solution. It put the ray of light on the solution
Thanks once again.
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