How to put files in zip file in macro

[mmea]

Hi 

I Where can I add that in this macro?

[yabwon · Accepted solution]

Like that?

%macro prefix(code);

proc datasets;
delete tabel1-tabel5;
quit;

/*tabel 1*/

proc sql;
create table tabel1 as select 
datetime() as dateformat ymdhms.,
lab,
ID,
from test
quit;

/*tabel2*/

proc sql;
create table tabel2 as select date format ymd., 
from test2
quit;

/*tabel3*/

proc sql;
create table tabel3 as select 
date format ymdhms.,
ID, 
CODE,
TESTED
from test3
quit;

/*tabel 4*/

data tabel4;
set test4;
format Dato ymd.;
run;

/*tabel 5 */

data tabel5;
set test5;
run;


%do i=1 %to 5;

proc export data=tabel&i. replace
  outfile="&out_folder.\&code._COVID19_TABLE&i._&YYYYMMDD..csv"
  dbms=dlm
;
delimiter=';';
quit;

/* try this ****/
filename in "&out_folder.\&code._COVID19_TABLE&i._&YYYYMMDD..csv" lrecl=1 recfm=n;
filename out "&out_folder.\&code..zip" member = "&code._COVID19_TABLE&i._&YYYYMMDD..csv" lrecl=1 recfm=n;

data _null_;
  rc = fcopy("in", "out");
  if rc ne 0 then 
    put "ERROR: for file &code._COVID19_TABLE&i._&YYYYMMDD..csv";
run;

filename in clear;
filename out clear;
/***************/
%end;
%mend;

%macro dummy_macro();
%if &nobs ne 4 %then %do; 

filename mymail email from='test@test.dk' to=&mailto.
         subject='ERROR -'
         attach=("&logpath.&YYYYMMDD._log.log");

%end;
%else %do;

%prefix(1);
%prefix(2);
%prefix(3);
%prefix(4);
%prefix(5);

filename mymail email from='test@test.dk' to=&mailto.
         subject='success';

ods listing file="&out_folder.\tesxtfile.txt";
data _null_;
    file print;
    text="The files has been uploaded succesfully";
    put text;
run;
ods listing close;


%end;
%mend;

%dummy_macro();

Bart

[mmea]

It says 

ERROR 23-2: Invalid option name member.

[yabwon]

Sorry, should be:

filename out ZIP "&out_folder.\&code..zip" member = "&code._COVID19_TABLE&i._&YYYYMMDD..csv" lrecl=1 recfm=n;

B

[mmea]

Thank you it worked - in my folder I willl get the CSV files seperate from the ZIP files - is there any way to ONLY have ZIP files in the folder?

[yabwon]

try this:

/* try this */
filename out ZIP "&out_folder.\&code..zip" member = "&code._COVID19_TABLE&i._&YYYYMMDD..csv" lrecl=1 recfm=n;

proc export data=tabel&i. replace
  outfile=out
  dbms=dlm
;
delimiter=';';
quit;

and drop my code between:

/* try this ****/

and

/***************/

If it fails, the try to replace data step from my code with:

data _null_;
  rc = fcopy("in", "out");
  if rc ne 0 then 
    put "ERROR: for file &code._COVID19_TABLE&i._&YYYYMMDD..csv";
  else fdelete("in");
run;

All the best

Bart