Re: Get list of first level subdirectories

Simple first level subdirectories list:

my $dir  = shift || "client";
my @list = ();

for( glob "*$dir*" ){
    -d or next;
    for( glob "$_/*" ){
        -d or next;
        push @list, $_;
    }
}

$" = $/;

print "@list"

__END__
client1/DIR_1
client1/DIR_2
client1/DIR_3
client2/DIR_1
client2/DIR_2
client2/DIR_3
client3/DIR_1
client3/DIR_2
client3/DIR_3
[download]

Comment on Re: Get list of first level subdirectories Download Code

Replies are listed 'Best First'.
Re^2: Get list of first level subdirectories by itub (Priest) on Jul 21, 2005 at 15:25 UTC
You don't need the outer loop, because `glob` can take care of that implicitly. And you can also use `grep` instead of `for` to build the list: `my $dir = shift \|\| "client"; my @list = grep -d, glob "$dir/*"; $" = $/; print "@list"` [download]	[reply] [d/l] [select]
Re^3: Get list of first level subdirectories by holli (Abbot) on Jul 21, 2005 at 15:57 UTC
`# 1 2 3 #123456789012345678901234567890123 print join"\n",grep-d,glob(shift)` [download] 33 :) holli, /regexed monk/	[reply] [d/l]
Re^4: Get list of first level subdirectories by sh1tn (Priest) on Jul 21, 2005 at 20:03 UTC
`-d&&print$_,$/for+glob+shift 29` [download]	[reply] [d/l]
Re^4: Get list of first level subdirectories by itub (Priest) on Jul 21, 2005 at 16:56 UTC
I knew someone wouldn't resist the temptation to golf it! ;-) But that wasn't my intention; I just wanted to make the code reasonably concise and clear (according to my definition of clear), and I also assumed that @list would be needed later in the code.	[reply]