in reply to Advanced Sorting

In that case, the only optimization to your code I can think of is: 
if ($field eq "name") {
  $a->{$field} cmp $b->{$field}
}
else {
  $a->{$field} <=> $b->{$field}
}

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which stops the code from examining every value twice, but simply evaluates a scalar. It benchmarks as slightly faster. Actually, for ultimate speed, move the field check before the sort subroutine is called: 
if ($field eq "name") { @sorted = sort @data; }
else { @sorted = sort {$a <=> $b} @data; }

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etc...

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RE: Re: Advanced Sorting
by Anonymous Monk on Oct 28, 2000 at 06:02 UTC
    I am just getting into perl,and have a basic simple question. I have written a script that reads in a directory of logfiles, sorts them in reverse chronological order, and outputs them into an html page. I cannot figure out how to sort the directory deeper than just the month, which comes first (the files in the dir are all listed by month/day/year). How can I sort this so the most recent days are first?
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      1. read the perlman (or unixman) for the commands stat() and sort()
      2. try to understand the following :-) 
      sub bydate{(stat $dir.$a)[9]<=>(stat $dir.$b)[9];}

@sortedfiles = sort bydate @files_in_your_dir;

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      Have a nice day
      All decision is left to your taste
      Update
      ok, I might have been wrong when assuming that logs get analyzed but not changed, after they have been archived
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[d/l]
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