perl folder list with "..", without ".".

Shell Programming and Scripting
1

Hi Everyone,

if my folder "foldera" inside has one file.
so if i do

 if ($df =~ /^\./) { next; }

then i will get
###
filea
###

if i want to have
###
..
filea
###
means also display the parent .., how should i modify the perl ~// in my code?

Thanks

---------- Post updated at 06:49 AM ---------- Previous update was at 03:15 AM ----------

I figure out one way, i know it is not good, but it works.

if ($df =~ /^\.\./) {
This is Parent Directory
} elsif ($df =~ /^\./) {
} else {
This is files.
}
2

Or, simpler:

next if $df=~/^\.$/;
3

Thanks :slight_smile:

4

if you read a list of files into an array, then '.' is the first entry assuming nothing else is going on that I am not aware of. So just skip the first element of the array when displaying the file/folders:

open DIR, '.';
my @files = readdir(DIR);
for my $i (1..$#files) {
   print "$files[$i]\n";
}

This should be more efficient than testing every element of the array with a regexp. But if you really need to test then use a string comparison operator instead of a regexp:

next if $df eq '.';

You don't need a regexp because you are not looking for a pattern.

5

Hi Kevin,
You mentioned "then '.' is the first entry assuming nothing else is going on that I am not aware of. So just skip the first element of the array when displaying the file/folders:", but sometimes let's say i have few files in a folder, when you print the file name, you may not always get the "." in the first, and ".." in the second line. So i may use sort first.

open DIR, '.';
my @files = readdir(DIR);
@new_files=sort @files; 
for my $i (1..$#new_files) {
   print "$files[$i]\n";
}

Do not know whether i thought correct or not. :eek:

6

If you sort the array first the . and .. could end up at other positions in the array, but when I did a test only three ASCII characters sort before . and ..

+
,
-

but your mileage could vary

You could try this:

open DIR, '.';
my @files = readdir(DIR);
@new_files=sort @files[1 .. $#files]; #leave out the first element of the array
for (@new_files) {
   print "$_\n";
}
7 
ls -a | egrep -v '^\.$'