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Date:   Tue, 19 Aug 2008 08:41:26 -0500
Reply-To:   Robin R High <rhigh@UNMC.EDU>
Sender:   "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:   Robin R High <rhigh@UNMC.EDU>
Subject:   Re: how to convert char variables into a date format?
Comments:   To: n <nikhil.abhyankar@GMAIL.COM>
In-Reply-To:   <ede98ea6-db24-4620-891a-59a8ccf45150@a70g2000hsh.googlegroups.com>
Content-Type:   text/plain; charset="US-ASCII"

The ANYDTDTE function comes in handy here as well as situations with other character date variables:

date

09-Apr-07 09/Apr/08 09/Apr/08 Apr08 04092008 20080409

_date=input(date,ANYDTDTE11.); format _date mmddyy10.; * assign date format of your choice;

date _date

09-Apr-07 04/09/2007 09/Apr/08 04/09/2008 09/Apr/08 04/09/2008 Apr08 04/01/2008 04092008 04/09/2008 20080409 04/09/2008

Robin High UNMC

n <nikhil.abhyankar@GMAIL.COM> Sent by: "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU> 08/19/2008 01:26 AM Please respond to n <nikhil.abhyankar@GMAIL.COM>

To SAS-L@LISTSERV.UGA.EDU cc

Subject how to convert char variables into a date format?

Hi all;

I have a char9 variable which actually holds dates like, 09-Apr-07.

I get an error, The informat $DATE was not found or could not be loaded. when I try using,

data want; set have; informat date DATE7. ; run;

Could anybody please tell me what is going wrong and how could I change the informat of the char type variable to a date informat?


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