topic Re: pulling today's date from group of files in SAS Enterprise Guide

pulling today's date from group of files

woo — Sat, 11 Apr 2015 00:10:46 GMT

Hi Friends please help me to pull today's date from below group of files

I have too many group file like below in one directory and i want to pull only today's file

Dir: d:\woo\test_dir

files under this dir are like;

customer_data_2015.04.06_test.txt

customer_data_2015.04.07_test.txt

customer_data_2015.04.08_test.txt

customer_data_2015.04.09_test.txt

customer_data_2015.04.10_test.txt

product_2015.04.08_test.txt

product_2015.04.09_test.txt

product_2015.04.10_test.txt

plant_detail_description_2015.04.04_test.txt

plant_detail_description_2015.04.05_test.txt

plant_detail_description_2015.04.06_test.txt

plant_detail_description_2015.04.07_test.txt

plant_detail_description_2015.04.08_test.txt

plant_detail_description_2015.04.09_test.txt

plant_detail_description_2015.04.10_test.txt

so for example i just need latest files (lets say today's file);

customer_data_2015.04.10_test.txt

product_2015.04.10_test.txt

plant_detail_description_2015.04.10_test.txt

Re: pulling today's date from group of files

art297 — Sat, 11 Apr 2015 00:58:02 GMT

My perl skills are at an extremely novice level, but I think that the following example achieves what you want:

data have;

informat directory $80.;

input directory;

cards;

customer_data_2015.04.06_test.txt

customer_data_2015.04.07_test.txt

customer_data_2015.04.08_test.txt

customer_data_2015.04.09_test.txt

customer_data_2015.04.10_test.txt

product_2015.04.08_test.txt

product_2015.04.09_test.txt

product_2015.04.10_test.txt

plant_detail_description_2015.04.04_test.txt

plant_detail_description_2015.04.05_test.txt

plant_detail_description_2015.04.06_test.txt

plant_detail_description_2015.04.07_test.txt

plant_detail_description_2015.04.08_test.txt

plant_detail_description_2015.04.09_test.txt

plant_detail_description_2015.04.10_test.txt

;

data want;

set have;

format y date9.;

re=prxparse('/\d\d\d\d\.\d\d\.\d\d/');

x=prxmatch(re,directory);

if x gt 0 then do;

if input(substr(directory,x,10),yymmdd10.) eq today();

end;

run;

Re: pulling today's date from group of files

Jagadishkatam — Sat, 11 Apr 2015 00:59:05 GMT

data have;

input txt$100.;

date=input(substr(txt,prxmatch('/\d{4}\.\d{2}\.\d{2}/',txt),10),yymmdd10.);

cards;

customer_data_2015.04.06_test.txt

customer_data_2015.04.07_test.txt

customer_data_2015.04.08_test.txt

customer_data_2015.04.09_test.txt

customer_data_2015.04.10_test.txt

product_2015.04.08_test.txt

product_2015.04.09_test.txt

product_2015.04.10_test.txt

plant_detail_description_2015.04.04_test.txt

plant_detail_description_2015.04.05_test.txt

plant_detail_description_2015.04.06_test.txt

plant_detail_description_2015.04.07_test.txt

plant_detail_description_2015.04.08_test.txt

plant_detail_description_2015.04.09_test.txt

plant_detail_description_2015.04.10_test.txt

;

%let td=%sysfunc(today());

%put &td;

proc sort data=have out=want;

by txt;

where date=&td;

run;

Re: pulling today's date from group of files

art297 — Sat, 11 Apr 2015 01:07:30 GMT

Ok! If perl functions can be used that way:

data have;

informat directory $80.;

input directory;

cards;

customer_data_2015.04.06_test.txt

customer_data_2015.04.07_test.txt

customer_data_2015.04.08_test.txt

customer_data_2015.04.09_test.txt

customer_data_2015.04.10_test.txt

product_2015.04.08_test.txt

product_2015.04.09_test.txt

product_2015.04.10_test.txt

plant_detail_description_2015.04.04_test.txt

plant_detail_description_2015.04.05_test.txt

plant_detail_description_2015.04.06_test.txt

plant_detail_description_2015.04.07_test.txt

plant_detail_description_2015.04.08_test.txt

plant_detail_description_2015.04.09_test.txt

plant_detail_description_2015.04.10_test.txt

;

data want;

set have;

if input(substr(directory,

prxmatch('/\d\d\d\d\.\d\d\.\d\d/',directory),10),yymmdd10.)

eq today();

run;

Re: pulling today's date from group of files

jakarman — Sat, 11 Apr 2015 09:40:33 GMT

A pitty yuo are not using Unix the filenames pattern/wildcards could solve that more elegant: SAS(R) 9.4 Companion for UNIX Environments, Fourth Edition
Windows is only allowing the * but that one could be sufficient having "today-s date" as a macro var available. Based on the naming convention of your files not the system dates.

customer_data_2015.04.10_test.txt

product_2015.04.10_test.txt

plant_detail_description_2015.04.10_test.txt

Will become:

%let today=%sysfunc(putn("10Apr15"d, yymmddp10.) ) ; %put &today ; /* when needed the date as constant */

%let today=%sysfunc(putn("&sysdate"d, yymmddp10.) ) ; %put &today ; /* when needed as moving system date */

/* other methods as next date as result of previous are to be coded */

For use in the infile:

"customer_data_&today._test.txt"

"product_&today._test.txt"

"plant_detail_description_&today._test.txt"

An all files of today (use the filename option of the infile statement:

"*_&today._test.txt"

Re: pulling today's date from group of files

Reeza — Sat, 11 Apr 2015 17:21:16 GMT

I agree with Jaap, calculate the date you need and use that instead of searching the file names.

Re: pulling today's date from group of files

woo — Sat, 11 Apr 2015 21:51:30 GMT

Thanks All...i am going to try this out....

Re: pulling today's date from group of files

woo — Sat, 11 Apr 2015 23:47:17 GMT

i used "prxmatch" function is it worked fine - thanks all...