topic Re: SAS LIBNAME a directory containing folders with datasets in SAS Programming

SAS LIBNAME a directory containing folders with datasets

alex_liang — Mon, 23 Jul 2018 02:18:50 GMT

Hi,

I was wondering if I could libname a directory containing serveral folders with datasets I need.

The directory looks like this:

Directory_A
--Directory_B
----Dataset_1.sas7bdat
--Directory_C
----Dataset_2.sas7bdat
----Dataset_3.sas7bdat
--Directory_D
----Dataset_4.sas7bdat

Cheers.

Re: SAS LIBNAME a directory containing folders with datasets

Reeza — Mon, 23 Jul 2018 02:58:29 GMT

You can have multiple paths in a library but I don't know of a way to have a wildcard reference. You could crawl the parent directory and concatenate the paths into one and pass it to a libname function or statement. I don't have time to code that up, but perhaps someone else can. There are examples of finding paths/folder structures on here or search SAS 9.4 Macro Appendix for an example.

@alex_liang wrote:
Hi,

I was wondering if I could libname a directory containing serveral folders with datasets I need.

The directory looks like this:

Directory_A
--Directory_B
----Dataset_1.sas7bdat
--Directory_C
----Dataset_2.sas7bdat
----Dataset_3.sas7bdat
--Directory_D
----Dataset_4.sas7bdat

Cheers.

Re: SAS LIBNAME a directory containing folders with datasets

ballardw — Mon, 30 Jul 2018 15:24:33 GMT

If you do not have many folders involved easiest might be a macro variable to hold the main part of the path common to all such as:

%let path=C:\Directory_A\;
libname mylib ("&path.directory_B &path.directory_C &path.directory_D");

of course using appropriate syntax for your operating system and care for letter case if needed.

The last \ is just part of Windows paths. Alternatively if you like you could use

%let path=C:\Directory_A;
libname mylib ("&path.\directory_B &path.\directory_C &path.\directory_D");

The periods are necessary to indicate the end of the macro variable value. Do not insert spaces if not in the in actual path if not there.

&path.\directory_B resolves to "C:\Directory_A\directory_B" &path.\ directory_B would resolve to "C:\Directory_A\ directory_B". One of which is incorrect for your use.