Hi.
I have a code that makes 25 CSV files.
These files are based on 5 regions and each region has 5 files.
I want to create 5 zipfiles for each regions, which includes the 5 files for each region.
I use this code to export to csv file (the 25 files), i use it in a macro to get the name of the file, so i my folder i get 25 files looking like this:
A_COVID19_TAB1_20210317.csv
A_COVID19_TAB2_20210317.csv
A_COVID19_TAB3_20210317.csv
A_COVID19_TAB4_20210317.csv
A_COVID19_TAB5_20210317.csv
B_COVID19_TAB1_20210317.csv
B_COVID19_TAB2_20210317.csv
B_COVID19_TAB3_20210317.csv
B_COVID19_TAB4_20210317.csv
B_COVID19_TAB5_20210317.csv
....
same for regioncode C, D, E
I want to make 5 zip files for each code named:
A_COVID19_20210317.zip
B_COVID19_20210317.zip
C_COVID19_20210317.zip
D_COVID19_20210317.zip
E_COVID19_20210317.zip
%let YYYYMMDD=%sysfunc(inputn(&DDMMMYYYY_h,best12.), YYMMDDN.6);
%macro regioncode(region);
%do i=1 %to 5;
proc export data=tabel&i. replace
outfile="&out_folder.\®ion._COVID19_TAB&i._&YYYYMMDD..csv"
dbms=dlm
;
delimiter=';';
quit;
%regioncode(A);
%regioncode(B);
%regioncode(C);
%regioncode(D);
%regioncode(E);
%end;
%mend;
... View more