Rename a file with perl - perl

I have a file in a different folder I want to rename in perl, I was looking at a solution earlier that showed something like this:
#rename
for (<C:\\backup\\backup.rar>) {
my $file = $_;
my $new = $file . 'backup' . $ts . '.rar';
rename $file, $new or die "Error, can not rename $file as $new: $!";
}
however backup.rar is in a different folder, I did try putting "C:\backup\backup.rar" in the <> above, however I got the same error.
C:\Program Files\WinRAR>perl backup.pl
String found where operator expected at backup.pl line 35, near "$_ 'backup'"
(Missing operator before 'backup'?)
syntax error at backup.pl line 35, near "$_ 'backup'"
Execution of backup.pl aborted due to compilation errors.
I was using
# Get time
my #test = POSIX::strftime("%m-%d-%Y--%H-%M-%S\n", localtime);
print #test;
To get the current time, however I couldn't seem to get it to rename correctly.
What can I do to fix this? Please note I am doing this on a windows box.

Pay attention to the actual error message. Look at the line:
my $new = $_ 'backup'. #test .'.rar';
If you want to interpolate the contents of $_ and the array #test into a string like that, you need to use:
my $new = "${_}backup#test.rar";
but I have a hard time making sense of that.
Now, strftime returns a scalar. Why not use:
my $ts = POSIX::strftime("%m-%d-%Y--%H-%M-%S", localtime);
my $new = sprintf '%s%s%s.rar', $_, backup => $ts;
Incidentally, you might end up making your life much simpler if you put the time stamp first and formatted it as YYYYMMDD-HHMMSS so that there is no confusion about to which date 04-05-2010 refers.

The line
my $new = $_ 'backup'. #test .'.rar';
probably should read
my $new = $file . 'backup' . #test . '.rar';
(You were missing a concatenation operator, and it is clearer to use the named variable from the line before than reusing $_ there...)

I think you missed the string concat symbol . (the period) :
my $new = $_ 'backup'. #test .'.rar';
should be
my $new = $_ . 'backup' . #test . '.rar';

A slight side issue but you don't need
for (<C:\\backup\\backup.rar>) {
my $file = $_;
.....
}
The < > construct would be useful if you were expanding a wildcard but you are not.
Be thoughtful of future readers of this code (you in a year!) and write
my $file = 'C:\backup\backup.rar' ;
Note the single quotes which doen't expand backslashes.

Related

Perl - Rename image files in directory

I don't use perl a whole bunch. I have a list of image files that I need to be renamed with an incrementing counter.
images folder
image_1_0.jpg
image_1_1.jpg
image_2_0.jpg
image_2_1.jpg
image_3_0.jpg
image_3_1.jpg
image_3_2.jpg
image_4_0.jpg
image_5_0.jpg
image_5_1.jpg
image_5_2.jpg
image_5_3.jpg
image_5_4.jpg
image_5_5.jpg
output would be
1.jpg
2.jpg
3.jpg
4.jpg
5.jpg
6.jpg
7.jpg
8.jpg
9.jpg
10.jpg
11.jpg
12.jpg
13.jpg
14.jpg
15.jpg
What I currently have
my $dir = usr/local/bin/images
my counter = 0;
opendir (IMGDIR, "$dir") or die "Cannot open directory: $!";
my #files = readdir(IMGDIR);
foreach my $oldfile(#files){
(my $oldfileb = $oldfile =~ s/\.[^.]+$//; #get file without extention
my $newfile = $dir/"$counter".jpg;
rename ("$dir/$oldfileb", "dir/$newfile");counter++;
}
Trying to use it more Perl more, but could use some help with this. Error is giving at the counter portion of code
Start the script with
#! /usr/bin/perl
use warnings;
use strict;
See strict and warnings. Perl will protect you from the most common errors.
Strings must be enclosed in quotes and every statement should end with a semicolon if another statement follows it:
my $dir = 'usr/local/bin/images';
Variables must start with a sigil:
my $counter = 0;
Parentheses must be closed:
(my $oldfileb = $oldfile) =~ s/\.[^.]+$//;
Since Perl 5.14, you can also use the more readable /r modifier:
my $oldfileb = $oldfile =~ s/\.[^.]+$//r;
Note that you should skip files that don't look like image names (readdir will return . and .. on *nix, for example). You also might want to sort the files.
/ outside of quotes is division, . is concatenation.
my $newfile = "$dir/$counter.jpg";
You already included $dir in $newfile:
rename "$dir/$oldfileb", $newfile;
Check the return value of rename for errors.
rename "$dir/$oldfileb", $newfile or warn "Can't rename $oldfile: $!";

Display folders using perl script

I have the following code and my problem is that I cannot modify it in order to use the $file3 outside the for function
for ($i = 0; $i < scalar(#temp); $i++){
$path7 = 'path_to'.#temp[$i];
foreach $path ($path7){
opendir my ($dh3), $path7 or die $!;
while ( my $file3 = readdir $dh3 ) {
next if $file3 eq '.' or $file3 eq '..';
next unless -d catfile($path7, $file3);
print "$file3\n";
}
closedir ($dh3);
}
}
Your $file3 is lexical to the while loop because you declared it with my. If you want it to be available outside, declare it in a larger scope, i.e. outside the for.
my $file3; # here!
for ( ...) {
# ...
# ...
######### no my below
while ( $file3 = readdir $dh3 ) {
# ...
}
# ...
}
Remember that in Perl it's a good practice to declare variables in the smallest scope necessary.
Also note that outside the while loop it will start out being undef and after being done processing the while for the first time ($i is 0, $path is the value of $path7), $file3 will keep the value it had in the last round of the while loop until the next time the while loop starts. That is never, because your foreach's list only has one element, as $path7 is a scalar and not an array. In fact, there is no need for that foreach loop at all. Just use $path7 directly.
Confused with my explanation because of the variable names? Me too. Always pick meaningfull variable names, don't just append numbers. That makes it very hard to maintain. :)

Perl - A way to get only the first (.txt) filename from another directory without loading them all?

I have a directory that holds ~5000 2,400 sized .txt files.
I just want one filename from that directory; order does not matter.
The file will be processed and deleted.
This is not the scripts working directory.
The intention is:
to open that file,
read it,
do some stuff,
unlink it and then
loop to the next file.
My crude attempt does not check for only .txt files and also has to get all ~5000 filenames just for one filename. I am also possibly calling too many modules?
The Verify_Empty sub was intended to validate that there is a directory and there are files in it but, my attempts are failing so, here I am seeking assistance.
#!/usr/bin/perl -w
use strict;
use warnings;
use CGI;
use CGI ':standard';
print CGI::header();
use CGI::Carp qw(fatalsToBrowser warningsToBrowser);
###
use vars qw(#Files $TheFile $PathToFile);
my $ListFolder = CGI::param('openthisfolder');
Get_File($ListFolder);
###
sub Get_File{
$ListFolder = shift;
unless (Verify_Empty($ListFolder)) {
opendir(DIR,$ListFolder);
#Files = grep { $_ ne '.' && $_ ne '..' } readdir(DIR);
closedir(DIR);
foreach(#Files){
$TheFile = $_;
}
#### This is where I go off to process and unlink file (sub not here) ####
$PathToFile = $ListFolder.'/'.$TheFile;
OpenFileReadPrepare($PathToFile);
#### After unlinked, the OpenFileReadPrepare sub loops back to this script.
}
else {
print qq~No more files to process~;
exit;
}
exit;
}
####
sub Verify_Empty {
$ListFolder = shift;
opendir(DIR, $ListFolder) or die "Not a directory";
return scalar(grep { $_ ne "." && $_ ne ".." } readdir(DIR)) == 0;
closedir(DIR);
}
Obviously I am very new at this. This method seems quite "hungry"?
Seems like a lot to grab one filename and process it!
Guidance would be great!
EDIT -Latest Attempt
my $dir = '..';
my #files = glob "$dir/*.txt";
for (0..$#files){
$files[$_] =~ s/\.txt$//;
}
my $PathAndFile =$files[0].'.txt';
print qq~$PathAndFile~;
This "works" but, it still gets all the filenames. None of the examples here, so far, have worked for me. I guess I will live with this for today until I figure it out. Perhaps I will revisit and see if anyone came up with anything better.
You could loop using readdir inside while loop. In that way readdir won't return all files but give only one at the time,
# opendir(DIR, ...);
my $first_file = "";
while (my $file = readdir(DIR)) {
next if $file eq "." or $file eq "..";
$first_file = $file;
last;
}
print "$first_file\n"; # first file in directory
You're calling readdir in list context, which returns all of the directory entries. Call it in scalar context instead:
my $file;
while( my $entry = readdir DIR ) {
$file = $entry, last if $entry =~ /\.txt$/;
}
if ( defined $file ) {
print "found $file\n";
# process....
}
Additionally, you read the directory twice; once to see if it has any entries, then to process it. You don't really need to see if the directory is empty; you get that for free during the processing loop.
Unless I am greatly mistaken, what you want is just to iterate over the files in a directory, and all this about "first or last" and "order does not matter" and deleting files is just confusion about how to do this.
So, let me put it in a very simple way for you, and see if that actually does what you want:
my $directory = "somedir";
for my $file (<$directory/*.txt>) {
# do stuff with the files
}
The glob will do the same as a *nix shell would, it would list the files with the .txt extension. If you want to do further tests on the files inside the loop, that is perfectly fine.
The downside is keeping 5000 file names in memory, and also that if processing this file list takes time, there is a possibility that it conflicts with other processes that also access these files.
An alternative is to simply read the files with readdir in a while loop, such as mpapec mentioned in his answer. The benefit is that each time you read a new file name, the file will be there. Also, you won't have to keep a large list of file in memory.

Using Perl to rename files in a directory

I'd like to take a directory and for all email (*.msg) files, remove the 'RE ' at the beginning. I have the following code but the rename fails.
opendir(DIR, 'emails') or die "Cannot open directory";
#files = readdir(DIR);
closedir(DIR);
for (#files){
next if $_ !~ m/^RE .+msg$/;
$old = $_;
s/RE //;
rename($old, $_) or print "Error renaming: $old\n";
}
If your ./emails directory contains these files:
1.msg
2.msg
3.msg
then your #files will look something like ('.', '..', '1.msg', '2.msg', '3.msg') but your rename wants names like 'emails/1.msg', 'emails/2.msg', etc. So you can chdir before renaming:
chdir('emails');
for (#files) {
#...
}
You'd probably want to check the chdir return value too.
Or add the directory names yourself:
rename('emails/' . $old, 'emails/' . $_) or print "Error renaming $old: $!\n";
# or rename("emails/$old", "emails/$_") if you like string interpolation
# or you could use map if you like map
You might want to combine your directory reading and filtering using grep:
my #files = grep { /^RE .+msg$/ } readdir(DIR);
or even this:
opendir(DIR, 'emails') or die "Cannot open directory";
for (grep { /^RE .+msg$/ } readdir(DIR)) {
(my $new = $_) =~ s/^RE //;
rename("emails/$_", "emails/$new") or print "Error renaming $_ to $new: $!\n";
}
closedir(DIR);
You seem to be assuming glob-like behavior rather than than readdir-like behavior.
The underlying readdir system call returns just the filenames within the directory, and will include two entries . and ... This carries through to the readdir function in Perl, just to give a bit more detail on mu's answer.
Alternately, there's not much point to using readdir if you're collecting all the results in an array anyways.
#files = glob('emails/*');
As already mentioned, your script fails because of the path you expect and the script uses are not the same.
I would suggest a more transparent usage. Hardcoding a directory is not a good idea, IMO. As I learned one day when I made a script to alter some original files, with the hardcoded path, and a colleague of mine thought this would be a nice script to borrow to alter his copies. Ooops!
Usage:
perl script.pl "^RE " *.msg
i.e. regex, then a file glob list, where the path is denoted in relation to the script, e.g. *.msg, emails/*.msg or even /home/pat/emails/*.msg /home/foo/*.msg. (multiple globs possible)
Using the absolute paths will leave the user with no doubt as to which files he'll be affecting, and it will also make the script reusable.
Code:
use strict;
use warnings;
use v5.10;
use File::Copy qw(move);
my $rx = shift; # e.g. "^RE "
if ($ENV{OS} =~ /^Windows/) { # Patch for Windows' lack of shell globbing
#ARGV = map glob, #ARGV;
}
for (#ARGV) {
if (/$rx/) {
my $new = s/$rx//r; # Using non-destructive substitution
say "Moving $_ to $new ...";
move($_, $new) or die $!;
}
}
I don't know if the regex fits the specifig name of the files, but in one line this could be done with:
perl -E'for (</path/to/emails*.*>){ ($new = $_) =~ s/(^RE)(.*$)/$2/; say $_." -> ".$new}
(say ... is nice for testing, just replace it with rename $_,$new or rename($_,$new) )
<*.*> read every file in the current directory
($new = $_) =~ saves the following substitution in $new and leaves $_ as intact
(^RE) save this match in $1 (optional) and just match files with "RE" at the beginning
(.*$) save everything until and including the end ($) of the line -> into $2
substitute the match with the string in$2

How can I separate a full path into directory and filename?

$a = '/etc/init/tree/errrocodr/a.txt'
I want to extract /etc/init/tree/errrocodr/ to $dir and a.txt to $file. How can I do that?
(Editor's note: the original question presumed that you needed a regular expression for that.)
Just use Basename:
use File::Basename;
$fullspec = "/etc/init/tree/errrocodr/a.txt";
my($file, $dir, $ext) = fileparse($fullspec);
print "Directory: " . $dir . "\n";
print "File: " . $file . "\n";
print "Suffix: " . $ext . "\n\n";
my($file, $dir, $ext) = fileparse($fullspec, qr/\.[^.]*/);
print "Directory: " . $dir . "\n";
print "File: " . $file . "\n";
print "Suffix: " . $ext . "\n";
You can see this returning the results you requested but it's also capable of capturing the extensions as well (in the latter section above):
Directory: /etc/init/tree/errrocodr/
File: a.txt
Suffix:
Directory: /etc/init/tree/errrocodr/
File: a
Suffix: .txt
you don't need a regex for this, you can use dirname():
use File::Basename;
my $dir = dirname($a)
however this regex will work:
my $dir = $a
$dir =~ s/(.*)\/.*$/$1/
I think a regex solution is a perfectly legitimate need - since my googling for exactly that brought me here. I want to pull out a filename in a group that's part of a larger match expression - here's my attempt:
~> echo "/a/b/c/d" | perl -ne '/(?<dir>\/(\w+\/)*)(?<file>\w+)/ && print "dir $+{dir} file $+{file}\n";'
dir /a/b/c/ file d
Use #array = split("/", $a);
and then $array[-1] is your file name.
For example:
$a =~ m#^(.*?)([^/]*)$#;
($dir,$file) = ($1,$2);
But, as other said, it's better to just use Basename for this.
And, BTW, better avoid $a and $b as variable names, as they have a special meaning, for sort function.
try this; it works at least for the filename, but when you modify it, it also gives you the direction:
You can also modify it on UNIX-systems for the \ instead if / or to use them both with |
$filename =~ s/(^.*/)//g;
somehow the backslash before the 2nd / is not displayed...
Maybe this will work:
#^(.+/)/([^/]+)$#