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Re: Checking for Directories
by btrott (Parson) on Jul 28, 2000 at 23:22 UTC
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You can check if a directory exists using -e, and if it's
a directory using -d. So you can probably just use the
latter:
print "is directory" if -d "/foo/bar";
For your second question:
opendir DH, "/foo/bar" or die $!;
my @subdirs = grep -d, readdir DH;
closedir DH or die $!;
Sub-directories are in @subdirs.
Umm, embarrassing update: read chip's answer
below for why my second answer is broken. So use his
code. | [reply]
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Sorry, btrott, that won't work ... readdir() only returns
basenames, and that's not enough for -d to work.
Besides, you probably don't want to count '.' and '..'. How about:
my $dir = '/foo/bar';
opendir DH, $dir or die $!;
my @subdirs = grep { -d } map { "$dir/$_" }
grep { $_ ne '.' && $_ ne '..' } readdir DH;
closedir DH;
-- Chip Salzenberg, Free-Floating Agent of Chaos
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#Assume all the open code
foreach( readdir DH ) {
next if ( $_ eq '.' or $_ eq '..' );
push @subdirs, "$dir/$_" if ( -d "$dir/$_" );
}
Of course, you may still not like this. It will be kinda
hungry on memory, especially if the ratio of files to
directories is very large - the foreach will suck in every
file at once. It may be better to try it with a while loop
- it will be easier on memory but will take longer due to
buffering issues.
mikfire | [reply]
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Oops. Thanks for the reminder. My test script was running
in the same directory that I was opendir-ing. Stupid,
but oh well. :)
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Or you could try something like this.
my @n = grep { -d $_ } <*>;
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Don't Check for Directories
by chip (Curate) on Jul 29, 2000 at 00:32 UTC
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I just noticed your purpose. I recommend that if you want to
create a directory if it's missing, just go ahead and call the
mkdir operator every time the program runs. The mkdir
may fail if the directory is there, but who cares? It's not like
there's any penalty for setting $!. :-)
-- Chip Salzenberg, Free-Floating Agent of Chaos
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Ah, but the sticky bit there is that you want to ignore
a failure due to "directory already exists" and not ignore
any other reasons for failure, including "file already
exists". Before Errno.pm, this was tricky. Now it isn't
so bad except that "file already exists" and "directory
already exists" put the same value in $!.
So I think the easiest robust solution is still to
only create the directory if it doesn't already exists.
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Well, I still don't see the point in checking ... after all,
what's the next thing you'll do? That's right, create a file
in the directory ... and whatever the error is, you can report
it.
Besides, why introduce a race condition without having to?
-- Chip Salzenberg, Free-Floating Agent of Chaos
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RE: Checking for Directories
by eLore (Hermit) on Jul 28, 2000 at 23:24 UTC
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One can use the -d "dirname" test.
from the command line:
perl -e "-d (\"/tmp\") and print \"aack\n\"";
in a script:
#!/usr/local/bin/perl
-d "/tmp" and print "/tmp is a directory\n";
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Just for future reference, you can mix quotation mark styles on the command line, eliminating the escapes.
$ perl -e "-d (\"/tmp\") and print \"aack\n\""
becomes
$ perl -e '-d ("/tmp") and print "aack\n"'
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Re: Checking for Directories
by Anonymous Monk on Jul 29, 2000 at 00:23 UTC
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Given a list of directories to be checked or made, this will
return a list of directories made. (@dirs NOT @created) is
the list of directories which already exist. If you @dirs
don't have a full path, use chdir.
my @created = grep { (!-d) && mkdir($_,oct(777)) } @dirs;
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RE: Checking for Directories
by Anonymous Monk on Jul 28, 2000 at 23:53 UTC
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-x $file does it exist
foreach your way through a directory handel testing w/ (-x $file && -d $file)
to test for subdirectories ... pop em ont an aray or whatever to store the vals | [reply] |