datetime function in sas mainframe

[rohitkrishna]

Hi all,

i'm facing sas DateTime function issue in my current job please find below the code 

{

     DATA _TIME;
INFILE DATALINES;
INPUT CMS_DAT $char30.;

CMS_DT = PUT(CMS_DAT, DATETIME30.6);
DATALINES;
22MAR2018:12:32:43.491538
06DEC2017:13:31:54.664700
03JAN2018:08:33:57.142454
17OCT2017:14:33:19.866942
28MAR2018:07:33:28.239593
; 

run;

and my rec is : 2018-03-22-12.32.43.491538 

}

so kindly sugeust solution for above sas mainframe code 

& if i remove that put statment i'm getting sum other values after mention put statment it gving exact input values i'm not getting expecting results has mention on the colour 

thanks & regards 

rohit

[Kurt_Bremser]

Use the proper informat and format:

data _time;
infile datalines;
input cms_dat datetime26.;
format cms_dat datetime26.6;
cms_dt = put(cms_dat, datetime30.6);
datalines;
22MAR2018:12:32:43.491538
06DEC2017:13:31:54.664700
03JAN2018:08:33:57.142454
17OCT2017:14:33:19.866942
28MAR2018:07:33:28.239593
; 

datetime values are numeric, read and handle them as such.

[rohitkrishna]

Hi Kurtbremser
thanks for the quick reply
{

CMS_DAT CMS_DT

22MAR2018:12:32:43.491538 22MAR2018:12:32:43.491538
06DEC2017:13:31:54.664700 06DEC2017:13:31:54.664700
03JAN2018:08:33:57.142454 03JAN2018:08:33:57.142454
17OCT2017:14:33:19.866942 17OCT2017:14:33:19.866942
. .
28MAR2018:07:33:28.239593 28MAR2018:07:33:28.239593

}

I'm getting above results are not exact results

i want my results should like: 2018-03-22-12.32.43.491538

thanks & regards
rohit

[rohitkrishna]

hi all
try to help out for the above problem
i want my results should be like

(2018-03-22-12.32.43.491538) I need output value,
not like
(22MAR2018:12:32:43.491538) this is an input value
thanks & regards
rohit

[rohitkrishna]

Hi all,

I required sum suggestion or solution for the below sas mainframe DateTime function job

(

DATA _TIME;
INFILE DATALINES;
INPUT CMS_DAT DATETIME26.;
FORMAT CMS_DAT DATETIME26.6;
CMS_DT = PUT(CMS_DAT, DATETIME30.6);
DATALINES;
22MAR2018:12:32:43.491538
06DEC2017:13:31:54.664700
03JAN2018:08:33:57.142454
17OCT2017:14:33:19.866942
28MAR2018:07:33:28.239593
;
RUN;

                                                                                }

and I want my results should be like (2018-03-22-12:32:43.491538) expected results 

try to suggest some solution for the above sas code 

thanks & regards 

rohit

[rohitkrishna]

sorry results should be like
2018-03-22-12.32.43.491538

[PaigeMiller]

You could concatenate strings created by the YYMMDDx. format and the TIME. format.

[Kurt_Bremser]

And here's an example of building a custom datetime format:

proc format library=work;
picture mydt
  low-high = '%Y-%0m-%0d-%0H.%0M.%0s' (datatype=datetime)
;
run;

data _time;
infile datalines;
input cms_dat datetime26.;
format cms_dat mydt26.6;
cms_dt = put(cms_dat,mydt26.6);
datalines;
22MAR2018:12:32:43.491538
06DEC2017:13:31:54.664700
03JAN2018:08:33:57.142454
17OCT2017:14:33:19.866942
28MAR2018:07:33:28.239593
;

proc print data=_time noobs;
run;

Result:

cms_dat                                 cms_dt

2018-03-22-12.32.43.491538    2018-03-22-12.32.43.491538
2017-12-06-13.31.54.664700    2017-12-06-13.31.54.664700
2018-01-03-08.33.57.142454    2018-01-03-08.33.57.142454
2017-10-17-14.33.19.866942    2017-10-17-14.33.19.866942
2018-03-28-07.33.28.239593    2018-03-28-07.33.28.239593

[Kurt_Bremser]

Please don't start multiple threads for the same question.

Since all standard formats use colons as separators in the time part, you will have to create a custom picture format with proc format.

Personally, I'd rather use the ISO 8601 international standard formats.

[rohitkrishna]

hi kurtbremser
I'm not getting you're point what you mention above statement
my requirement is to change the month
Jan to 01
Feb to 02 like that
remaining I can translate: to. I know that part but how to change this month is sum difficult to me
i use if condition for month change
if month = 'jan' then month = 01 like that
thanks & regards
rohit

[Kurt_Bremser]

We've already covered reading datetime strings with three-letter month names. To extract the month from the datetime value, use the datepart() and month() functions.

[PaigeMiller]

As implied by @Kurt_Bremser, you will not get where you want to go (or you will work very hard at it) unless you use the built-in SAS date and time functions (such as datepart and timepart) and date and time formats.

Here is an example that gets you the desired end result

data want;
	set _time;
	result = cats(put(datepart(cms_dat),yymmddd10.),'-',put(timepart(cms_dat),time18.6));
run;

[rohitkrishna]

Hi KurtBremser, PaigeMiller
Thanks for the help it superb it's working
Thanks & regards
rohit