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Gil_
Quartz | Level 8
I have a col classified character $31
Here is example.
2018-04-03:11:26:45:573000000

I use
Data test;
Set test;
Day=substr(left(opendate),1,10);
Format day mmddyy10.;
Run;

I used that b4 the input to convert it.
1 ACCEPTED SOLUTION

Accepted Solutions
SuryaKiran
Meteorite | Level 14
data want;
   set have;
   day=INPUT(SUBSTR(left(opendate),1,10),yymmdd10.);
   format day mmddyy10.;
   run;
Thanks,
Suryakiran

View solution in original post

6 REPLIES 6
Reeza
Super User

INPUT() converts and you can try ANYDTDTM. for a datetime

 


@Gil_ wrote:
I have a col classified character $31
Here is example.
2018-04-03:11:26:45:573000000

I use
Data test;
Set test;
Day=substr(left(opendate),1,10);
Format day mmddyy10.;
Run;

I used that b4 the input to convert it.

 

Gil_
Quartz | Level 8
It converted to number here is code
Data test;
Set test;
Day_=input(day,ANYDTDTM.);
FORMAT DAY_ ANTDTDTM.;
RUN;

The output is this
1838332800
I would need it to look like 04/04/2018
Reeza
Super User

 You didn't apply the format correctly and since it's a date time, it creates a date time. You can display it as a date using DTDDMMYY10 or DTDATE9 or convert it using DATEPART() to get the date portion.


@Gil_ wrote:
It converted to number here is code
Data test;
Set test;
DayTime = input(day,ANYDTDTM.);
date = datepart(dayTime);

FORMAT DAY dtdate9. date ddmmyy10.;
RUN;

SuryaKiran
Meteorite | Level 14
data want;
   set have;
   day=INPUT(SUBSTR(left(opendate),1,10),yymmdd10.);
   format day mmddyy10.;
   run;
Thanks,
Suryakiran
Astounding
PROC Star

Even simpler (the FORMAT statement is still necessary):

 

day = input(opendate, yymmdd10.);

 

That assumes your character variable OPENDATE is left-hand-justified.

Gil_
Quartz | Level 8
Thank you

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