topic Re: Date and Time Issue in SAS Programming

Date and Time Issue

Crubal — Fri, 09 Mar 2018 19:32:08 GMT

Hi,

I get trouble in date-time change issue. I would like to grab month and date information from datetime.

I would like to transfer table a, column 'End_Date' from '15MAR2001:00:00:00' to a new column 'End_Date_1': '15MAR' or '03-15'.

How could I do it?

Similarly, if I have a table b, column name 'Start_Date', from '2016-03-09' to a new column 'Start_Date_1': '09MAR' or '03-15'.

I hope new column's data type is 'date'.

Is there any way to do that?

Thanks!

Re: Date and Time Issue

Kurt_Bremser — Fri, 09 Mar 2018 19:41:42 GMT

The datepart() function extracts the date value from a datetime. After that, you can use the month() function.

Re: Date and Time Issue

Crubal — Fri, 09 Mar 2018 19:51:46 GMT

Thanks @Kurt_Bremser

I used

month(datepart(End_Date)) * 100 + day(datepart(End_Date))

to obtain the new date. And the output format looks like '101' / '315' ...

Although looks not pretty, I could compare anyway.

Re: Date and Time Issue

Bhushan — Fri, 09 Mar 2018 20:08:08 GMT

data new;
date='19APR2018'D;
Format date date9.;
Date1=month(date);
Date2=day(date);
New=Strip(put(Date1,$5.))||"-"||Strip(put(Date2,$5.));
run;

Re: Date and Time Issue

ballardw — Fri, 09 Mar 2018 20:21:00 GMT

@Crubal wrote:

Thanks @Kurt_Bremser

I used

month(datepart(End_Date)) * 100 + day(datepart(End_Date))

to obtain the new date. And the output format looks like '101' / '315' ...

Although looks not pretty, I could compare anyway.

If you want to compare things it is likely best to leave them as date values instead of creating ugly character variables.

If you want to see the date in a specific format there are a large number of date formats supplied by SAS. For example:

data example;
   x='14MAR2018'd;
   format x mmddyyd5.;
run;

will display the date value 14Mar2018 as 03-14. The last d in the format says to place a dash (-) between the values of month, day and year. The 5 says to only display the first 5 characters of the result.

Re: Date and Time Issue

SuryaKiran — Fri, 09 Mar 2018 20:34:15 GMT

If you have datetime values then extract date values using datetime() and apply format DATE5.

Date5. format will show you only the first 5 places.

DATA HAVE;
format DATETIME_val DATE5.;
DATETIME_val=DATEPART(INPUT('15MAR2001:00:00:00',datetime24.));
run;

Re: Date and Time Issue

Crubal — Fri, 09 Mar 2018 23:08:45 GMT

@Bhushan

Will 'strip' function in your code return 'string' data type?

Thanks!

Re: Date and Time Issue

Crubal — Fri, 09 Mar 2018 23:12:48 GMT

Hi @SuryaKiran

I try to replace it with old column name 'End_Date':

DATA HAVE;
format DATETIME_val DATE5.;
DATETIME_val=DATEPART(INPUT(End_Date,datetime24.));
run;

It seems the new column 'DATETIME_val' is empty.

Re: Date and Time Issue

Crubal — Fri, 09 Mar 2018 23:16:07 GMT

This's great. Thank you.

Re: Date and Time Issue

ballardw — Fri, 09 Mar 2018 23:35:51 GMT

@Crubal wrote:

Hi @SuryaKiran

I try to replace it with old column name 'End_Date':
DATA HAVE;
format DATETIME_val DATE5.;
DATETIME_val=DATEPART(INPUT(End_Date,datetime24.));
run;
It seems the new column 'DATETIME_val' is empty.

NO set statement to use your existing data set. So END_DATE has no value. You should have noticed that in the "have" set.

Re: Date and Time Issue

Crubal — Sat, 10 Mar 2018 01:48:42 GMT

Yeah, even if I have that information, I mean. Thanks!

Re: Date and Time Issue

Bhushan — Sat, 10 Mar 2018 13:44:27 GMT

@Crubal Yes Strip is a char function however datatype will be char