topic Re: Date Conversion in SAS Programming

Date Conversion

tobyfarms — Thu, 28 Sep 2017 18:17:55 GMT

I know there is a simple solution but I am having trouble finding it here..So if someone know where the solution is, can you please just post the link.

I am trying to change the date format from '01JAN2017' to '0116'. The '01JAN2017' field is currently in numeric date9. format.

Thank you.

Re: Date Conversion

PaigeMiller — Thu, 28 Sep 2017 18:26:51 GMT

Change the format on this variable to MMDDYY4.

Re: Date Conversion

ballardw — Thu, 28 Sep 2017 19:18:31 GMT

@tobyfarms wrote:

I know there is a simple solution but I am having trouble finding it here..So if someone know where the solution is, can you please just post the link.

I am trying to change the date format from '01JAN2017' to '0116'. The '01JAN2017' field is currently in numeric date9. format.

Thank you.

Where does the 16 come from for 01Jan2017? If isn't a month number, a day number or any reasonable year value?

If you mean 0101 for 01Jan2017 then @PaigeMiller is correct.

Re: Date Conversion

tobyfarms — Thu, 28 Sep 2017 20:12:14 GMT

@ballardw Sorry that was a mistake. I meant to have '0117', capturing the monthYear.

Re: Date Conversion

PBsas — Thu, 28 Sep 2017 20:49:43 GMT

data want;
format date mmyyn4.;
date='01jan2017'd;
run;

Try this

Re: Date Conversion

mkeintz — Thu, 28 Sep 2017 21:15:56 GMT

How about 2-digit year first, then 2-digit month? If so, you can use the yymmdd format, truncated to 4 characters.

data _null_;

d='27sep2017'd;

put d=yymmdd4.;

run;

editted additon: I didn't see a MMYY format, but @PBsas did. So you can ignore the below:

If you really need MMYY, I don't see a MMYY format. But you can make your own, using the PICTURE statement and the DATATYPE option, as in:

proc format;
  picture myformat (default=4)
    low-high = '%0m%0y' (datatype=date);
run;

data _null_;
  d=today();
  put d=myformat. ;
run ;

Re: Date Conversion

Tom — Thu, 28 Sep 2017 23:10:20 GMT

Why would you want the value in MMYY order? That will not sort properly and would be very confusing. Why YYMM?

Do you want a string?

str = substr(put(date,ddmmyy6.),3,4);

A number?

numb = month(date)+100 + year(date)-2000 ;
format numb z4. ;

If a number do want to store the actual date and just attach a format?

proc format;
 picture mydt other='%0m%0y' (datatype=date);
run;
....
format date mydt4.;

Re: Date Conversion

ShiroAmada — Fri, 29 Sep 2017 01:04:46 GMT

Try this...

data _null_;
  today=today();
format today MMYYN4.;
PUT TODAY;
run;

Hope this helps.

Re: Date Conversion

tobyfarms — Fri, 29 Sep 2017 18:00:21 GMT

The Substr function worked for what I am trying to achieve in my coding. This populate the 2-digit month and 2 digit year from my current date format. Thanks @Tom