in reply to multiple dir. grep

Why would you need grep? Why not just use sort? The only use for grep I can envision for this problem is if you wanted to remove duplicates. Is that what you're asking?

the lowliest monk

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Re^2: multiple dir. grep
by Cockneyphil (Acolyte) on Apr 06, 2005 at 13:29 UTC
    note: I am fairly new to perl I have 2 dirs one contains files with sequential filenames ( 12345.dat, 12346.dat, 12347.dat etc..) the other contains similar filenames but with an alpha prefix (A23456.dat, F23457.dat, S23458.dat etc..) I need to create a list that contains the filenames of both dirs -with the newest filenames (higest sequential number regardless of prefix) from both at the top the code i am using (at the moment) for generating my list is: 
    opendir THEDIR, "$dirpath" || die "cant open $dirpath !";
@allfiles = grep /^(?:158|159)/,readdir THEDIR; 
closedir THEDIR;
foreach $file (sort { ($b) cmp ($a) } @allfiles) {
print "$file\n";}

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    please don't laugh..it took me a LONG time to get that to work. As you can see to make it all go a bit faster i have limited the list to only filenames that start with either 158 or 159. What I need to do is somehow get the contents of the other dir into this list whilst still keeping the newest filenames at the top. ps I'm not asking for a complete script just a few pointers as to how to go about doing it.
[reply]
[d/l]

      It's quicker to code it than to explain it: 

      use strict;

my @dirs = qw(dir1 dir2 dir3);

my @files;
for my $dir ( @dirs ) {
  opendir my $dh, $dir or die "opendir failed: $dir ($!)\n";
  push @files, grep -f $_, readdir $dh;
}

my @sorted =
  map $_->[ 0 ],
  sort { $b->[ 1 ] <=> $a->[ 1 ] }
  map [ $_, (/(\d+)/)[ 0 ] ],
  @files;

__END__

      [download]

      the lowliest monk

[reply]
[d/l]

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