Hi! I would like to seek your help to solve my question!
My existing variable format is Char "2018-08-31 00:00:00" and I want to convert it to SAS Date9. format. Please see my code as below:
Data Temp;
set T;
format AA date9.;
AA=input(char,MMDDYY10.);
run;
Really hope you can help me! Thanks a lot!
data want; char_date="2018-08-31 00:00:00"; num_date=input(substr(char_date,1,10),yymmdd10.); format num_date date9.; run;
data want; char_date="2018-08-31 00:00:00"; num_date=input(substr(char_date,1,10),yymmdd10.); format num_date date9.; run;
Actually, the informat "substrings" on its own:
data have;
input char :$20.;
cards;
2018-08-31 00:00:00
;
run;
data want;
set have;
format aa date9.;
aa = input(char,yymmdd10.);
run;
True, but you know my thoughts on implicit anything.
Here's an example of how to:
1. Read it in as date time
2. Extract date portion
3. Extract time portion
4. Keep as datetime and format as date to view
data have; x="2018-08-31 00:00:00"; *import and convert to date time variable; y_datetime=input(x, anydtdtm.); *get only the date part; y_date=datepart(y_datetime); *get only the time part; y_time=timepart(y_datetime); *keep as datetime and format as date; Y_formatted = Y_datetime; format y_datetime datetime. y_date date9. y_time time. y_formatted dtdate9.; run; proc print data=have;run;
SAS Output
Obs | x | y_datetime | y_date | y_time | Y_formatted |
---|---|---|---|---|---|
1 | 2018-08-31 00:00:00 | 31AUG18:00:00:00 | 31AUG2018 | 0:00:00 | 31AUG2018 |
@New_SAS_user76 wrote:
Hi! I would like to seek your help to solve my question!
My existing variable format is Char "2018-08-31 00:00:00" and I want to convert it to SAS Date9. format. Please see my code as below:
Data Temp;
set T;
format AA date9.;
AA=input(char,MMDDYY10.);
run;
Really hope you can help me! Thanks a lot!
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