Accessing Default directory in perl

Question

I am having a perl script which is taking a directory path as input and then script iterates through all the files present in that directory. For each file it calls some other function.

Here is the script:

#!/usr/bin/perl

use JSON;   
use XML::Simple;
use File::Spec;

$num_args = $#ARGV + 1;
if ($num_args != 1) {
    print "\nUsage: $0 <input directory>\n";
    exit;
}

my $dir = $ARGV[0];

opendir(DIR, $dir) or die "cannot open directory";

@docs = grep(/\.xml$/,readdir(DIR));

foreach $file (@docs) 
{
    my $abs_path = join("",$dir,$file);
    #print $abs_path;
    my $return_object = some_function_call($abs_path);
    print_output($return_object);
 }

 sub print_output
 {
    my $output_object = $_[0];
    print $output_object;
 }

My question here is when i am reading input directory, in case if does not exists how can make above script read from "Some Default Directory location". I tried adding some logic but did not work for me. This is my first perl script so please excuse me if this is too basic to ask.

I am working in LINUX environment.

[You should always use `use strict; use warnings;`](http://stackoverflow.com/q/8023959/725418) — TLP, Feb 13 '15 at 15:34

score 5 · Accepted Answer · answered Feb 13 '15 at 15:47

The first and most critical mistake you have made is to not use

use strict;
use warnings;

These two pragmas provide more help than any other two lines of code you could possibly add. Never write Perl code without them.

use JSON;   
use XML::Simple;
use File::Spec;

You do not use either of these modules, so for this code, they are redundant.

$num_args = $#ARGV + 1;
if ($num_args != 1) {
    print "\nUsage: $0 <input directory>\n";
    exit;
}

This can be written

if (@ARGV != 1) {
    print ...
    exit;
}

Because an array in scalar context returns its size.

opendir(DIR, $dir) or die "cannot open directory";
@docs = grep(/\.xml$/,readdir(DIR));
foreach $file (@docs) {
    my $abs_path = join("",$dir,$file);

Unless you really need the precision of readdir, what you most often want is glob(). This code will replace the above code:

my @docs = glob "$dir/*.xml";

It will return the relative path (not the absolute path, as your variable name implies -- unless the path argument is absolute). For example: examples/foo.xml.

sub print_output {
    my $output_object = $_[0];
    print $output_object;
}

I assume this sub is just an example, because it does nothing, and can be reduced to a simple print.

As to your question about default directory to read from, that to me is at odds with the requirement of an argument (where you check @ARGV). But you can do a simple check for existence:

if (not -e $dir) {
    $dir = "default";
}

yes i will make sure i will use use strict; use warnings; , in my perl scripts. This was first time i have been working on perl so did not knew much about the good perl programming practices. — user1188611, Feb 13 '15 at 18:58

score 2 · Answer 2 · answered Feb 13 '15 at 16:56

The constant pragma and a conditional statement modifier would make your default logic rather readable:

....
use constant DEFAULT_DIR => '/path/to/default/dir';
....

my $dir = $ARGV[0];
$dir = DEFAULT_DIR unless -e $dir;
....

(And, yes, do take notice of others' suggestions: strict, warnings, lexical dirhandles, pruning unneeded modules, avoiding $#array, etc.)

score 1 · Answer 3 · edited Feb 13 '15 at 15:43

1

Before opendir(DIR, $dir) or die "cannot open directory"; check if directory exists and if not assign $dir variable default value:

$default_dir = "....";

if (!-e $dir)
{
    $dir = $default_dir;
}

edited Feb 13 '15 at 15:43

TobyLL

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answered Feb 13 '15 at 15:31

Kostia Shiian

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Accessing Default directory in perl

3 Answers3

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