如何使用perl从给定目录中读取所有文件名？

Question

friends i 've this code which read file name from a directory and print those names

opendir DIR1,  "defaults" or die "cannot open dir: $!";#open the given dir
        my @default_files=readdir DIR1;
        foreach my $fls(@default_files){ 
                        chomp($fls); 
                        print "$fls\n";
           }

when i run the above code i've got

.
..
PGR.csv
report.csv
std_headers.csv
tab_name.txt

'm confused what is first two lines? thanks in advance

Answer 1

A simpler alternative can be using the glob function:

my @default_files = glob 'defaults/*';

Or:

my @default_files = <defaults/*>;

These functions filter out file and directory entries that begin with . , just like the shell does.

Answer 2

The first to lines are:

• . link to the directory you are currently visiting.

• .. link to the directory above the directory (parent directory) you are visiting.

EDIT:

One possibility to filter out both named entries would be using grep :

opendir DIR1, "tmp" or die "cannot open dir: $!";
my @default_files= grep { ! /^\.\.?$/ } readdir  DIR1;

foreach my $fls (@default_files) { 
   chomp($fls); 
   print "$fls\n";
}

Answer 3

"." means the present working directory and ".." means the parent directory of the present working directory

Answer 4

In a directory listing, . stands for Working Directory and .. stands for Parent Directory .

In scalar context, you can skip those entries by explicitly checking for them:

while (defined(my $next = readdir $dir)) {
    next if $next =~ /\A[.][.]?\z/;
    # ...
}

or in list context:

my @entries = grep !/\A[.][.]?\z/, readdir $dir;

If I am already using File::Slurp , I prefer to just use the read_dir function it provides. The following will automatically skip . and .. in addition to prepending the path to the directory to each returned entry:

 my @paths = read_dir('/path/to/dir', prefix => 1);