Filename should be same as specified date

[Kirito1]

First I was trying to give the filename the exact date on which I exported the file.

proc export data=AVG
outfile="xyz/Transactions \%sysfunc( today(), date9 ).csv"
REPLACE
DBMS=CSV;
run;

This is giving me an output filename as  (Transactions 22DEC2022).

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But now I want the date which I specify in a code should be attached with the File name.

For Example: I specified a date in code as 19DEC2022  and the file name is Transactions. So, the output filename should be Transactions 19DEC2022.

I think this problem could be solved using macros but I am not sure how to proceed it. Please! help.

[Kurt_Bremser]

From longtime IT experience, two tips:

• do not use DMY dates in filenames. Use YMD instead, so the files sort in chronological order on their own, and it is easiervto use wildcards to pick years or months.
• avoid blanks in filenames. Use underlines instead.

How do you "specify a date in a code"? Please show that particular piece of code.

Use the "little running man" button to open a window into which you paste the code. This keeps formatting, leaves the code as is (no HTML formatting), and provides coloring similar to the SAS Enhanced Editor.

[Kirito1]

That's the question how can I just name a file using the date I specify. Sorry I think this is a bit confusing but is there a way where we %let to specify date and then just like called today(), I could call a date that I want with the file name to attach.

[Kurt_Bremser]

To store today's date in a macro variable, do

%let today = %sysfunc(today(),yymmddn8.);

If you want to use the date in code (for calculations or comparisons), omit the format in %SYSFUNC.

[Kirito1]

But, what if I want to store yesterday's date. Then how can do that?

[Kurt_Bremser · Accepted solution]

SAS dates are counts of days, starting with 1960-01-01 as day zero.

So you only need to subtract a day from today:

%let yesterday = %sysfunc(putn(%sysfunc(today())-1,yymmddn8.));

Or you use a DATA step:

data _null_;
call symputx("yesterday",put(today()-1,yymmddn8.));
run;

[Kirito1]

Sorry to bug you so much, but what will we do if we want to store a date as crazy as 1999-10-24 with the file name while exporting it.

[Kurt_Bremser]

From where do you get the date? Is it calculated in code, read from a dataset, or supplied in another way to your code?

[Kirito1]

Supplied in another way, For example I am working on a task that was supposed to be submitted on so and so date but I am doing it today and want I want that the file name should consist the date when the file was supposed to be worked on.
Like I export a file with today() it will save file on my system with today's date but I want that file to be saved as 16-09-2022. Then I how could I do that.

[Kurt_Bremser]

Either you have a way to automatically calculate the date, or you need to write it literally into the code anyway.

I have already shown how to store a calculated date into a macro variable.

[Kirito1]

proc import datafile='xyz/IMPORT/TXN_DATA.csv'
out=CUSTAVG
DBMS=CSV
REPLACE;
run;
/*Defining yesterday as a function*/
%let yesterday = %sysfunc(putn(%sysfunc(today())-1,yymmddn8.));
/*Exporting a file*/
proc export data=CUSTAVG
outfile="xyz/Transactions \%sysfunc( yesterday(), date9 ).csv"
REPLACE
DBMS=CSV;
run;

Today() was working but if I use yesterday even after defining a macro Its giving me error.
Below Code was working fine and giving me desired results.
proc export data=AVG
outfile="xyz/Transactions \%sysfunc( today(), date9 ).csv"
REPLACE
DBMS=CSV;
run;

[Kurt_Bremser]

yesterday is a macro variable, not a data step function.

outfile="xyz/Transactions \&yesterday..csv"