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Same principle; using the lengthn function, the check for >=12 is there to account for names less than 5 long.
data set1;
informat name $200.;
infile cards;
input name;
if lengthn(name) >=12 then
name=substr(name,1,5) || substr(name,lengthn(name)-6,7);
cards;
Evangelina194511F
Ann194511F
;
run;
-Mary
----- Original Message -----
From: Irin later
To: SAS-L@LISTSERV.UGA.EDU ; Mary
Sent: Thursday, May 01, 2008 2:25 PM
Subject: Re: How to take out the part of the string ID? Need help.
Mary, it does work! Thank you so much!
I just realized that I have one more problem. Woul you be so kind as to help me with that?
I compare two files and it appears that id on the other side have longer name:
Evangelina194511F
How can I constract it in order to have just 5 letters of the name:
Evang194511F
Thank you for ALL YOUR HELP!
Irin
--- On Thu, 5/1/08, Mary <mlhoward@avalon.net> wrote:
From: Mary <mlhoward@avalon.net>
Subject: Re: How to take out the part of the string ID? Need help.
To: irinfigvam@yahoo.com, SAS-L@LISTSERV.UGA.EDU
Date: Thursday, May 1, 2008, 3:05 PM
data set1;
informat name $200.;
infile cards;
input name;
name=substr(name,1,lengthn(name)-3) || substr(name,lengthn(name),lengthn(name));
cards;
ADAM19480204M
;
run;
-Mary
----- Original Message -----
From: Irin later
To: SAS-L@LISTSERV.UGA.EDU
Sent: Thursday, May 01, 2008 1:15 PM
Subject: How to take out the part of the string ID? Need help.
I have newid for example
ADAM19480204M (name+year+month+day+gender)
I need to rebuild it in order it looks like:
ADAM194802M (name+year+month+gender)
(Name can vary up to 5 letters)
In other words I need to take out DAY part.
Could you give me a hand ,please , with this code?
Thank you in advance, Irin
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