glob files by number sequence (date)

[dlite922]

Hey Guys,

I'm a PHP guy, but choose to do my file scripts in perl and have come upon this problem:

I have n files with a date in their name: test2010-08-23.txt for example.

I want grab (using glob or whatever else) the last x number of files.

Example assuming today is Aug 23rd, the last 4 days would grab the files for the 22nd, 21st, 20th, and 19th of August.
test2010-08-19.txt
test2010-08-20.txt
test2010-08-21.txt
test2010-08-22.txt

Is there a better way to do this other than back tracing with a -1 forloop?

// my current plan in *pseudo* code, the forloop
 
days = 7
@testfiles = array(); 
 
for(i = days; i > 0; i--)
{
   aFile = 'test'.formatdate("yyyymmdd",today - i).'.txt'; 
   if(file_exists(aFile)
      push(@testfiles,aFile);
}
 
 
 

Just extending my perl knowledge.


Thanks All,


Dan

[Oralloy]

Probably not for the date selection.

You will have to deal with month and year boundaries, so I'd suggest that you use a date processing module, and not just ad-hoc code. In the long run you'll save yourself a whole lot of grief.

Also, does your specification say the last four files, regardless of date, or just the files from the last four days? Something to think about.

[dlite922]

Yes, I am thinking about those things, but wanted to keep the post simple.

I'm going to use Date:Calc for the date processing.

I will be ignoring files that the dates do not exist. (i.e. if 3 files exist for the 4 files, then i'll only take 3 files)

It is for an averaging functions so in the end i'll sum the numbers I need from each file and divide by the number of files (in the above case: divide by 3 instead of 4).

Thanks Though!

I'll figure something out tomorrow and if solution is related to OP i'll post the code here.


Cheers,


Dan