I have a folder called "Lib" in my drive it contains many files inside and I have a problem that this "Lib" folder is there in many other places in the drive. My Perl script has to copy the contents from folder "Lib" which are latest updated and paste it in the folder "d:\perl\Latest_copy_of_Lib"
For example, I have a Lib folders in d:\functions, d:\abc, and many other places. I want to find the latest copy of each file in those directories. So, if the file d:\functions\foo.txt was last modified on 2009-10-12 and d:\abc\foo.txt was last modified on 2009-10-13, then I want the version in d:\abc to by copied to the target directory.
I have used file::find but it searches in whole dir and copies the contents that are not latest copy.
I think you just described rsync. Unless you have some sort of weird requirements here, I don't think you need to write any code to do this. I certainly wouldn't reach for Perl to do the job you described.
You need to use File::Find to create a hash of files to move. Only put the path to a file in the hash if the file is newer than the path already stored in the hash. Here is a simple implementation. Note, there may be problems on the windows platform, I am not used to using File::Spec to work with files and pathes in a cross platform manner.
#!/usr/bin/perl
use warnings;
use strict;
use File::Find;
use File::Spec;
my %copy;
my #sources = qw{
/Users/cowens/foo/Lib
/Users/cowens/bar/Lib
/Users/cowens/baz/Lib
};
find sub {
my ($volume, $dir, $file) = File::Spec->splitpath($File::Find::name);
my #dirs = File::Spec->splitdir($dir);
my #base = ($volume); #the base directory of the file
for my $dir (#dirs) {
last if $dir eq 'Lib';
push #base, $dir;
}
#the part that is common among the various bases
my #rest = #dirs[$#base .. $#dirs];
my $base = File::Spec->catdir(#base);
my $rest = File::Spec->catfile(#rest, $file);
#if we don't have this file yet, or if the file is newer than the one
#we have
if (not exists $copy{$rest} or (stat $File::Find::name)[9] > $copy{$rest}{mtime}) {
$copy{$rest} = {
mtime => (stat _)[9],
base => $base
};
}
}, #sources;
print "copy\n";
for my $rest (sort keys %copy) {
print "\t$rest from $copy{$rest}{base}\n";
}
If you can standardize on a single location for your libraries, and then use one of the following:
set PERL5LIB Environment variable and add
use lib 'C:\Lib';
or
perl -I C:\Lib myscript
Any of these will give you a single copy of your lib directory that any of your scripts will be able to access.
Related
I'm trying to zip up image files using Archive::Zip. The files are in Data/Temp/Files When I loop through the logs in the directory and add them to the zip file, I end up with the folder hierarchy and the image files when I only want the image files.
So the zip ends up containing:
Data
└Temp
└Files
└Image1.jpg
Image2.jpg
Image3.jpg
When I want the zip file to contain is:
Image1.jpg
Image2.jpg
Image3.jpg
Here is the script I'm running to test with:
#!/usr/bin/perl
use Archive::Zip;
$obj = Archive::Zip->new(); # new instance
#files = <Data/Temp/Files/*>;
foreach $file (#files) {
$obj->addFile($file); # add files
}
$obj->writeToFileNamed("Data/Temp/Files/Images.zip");
Use chdir to change into the directory:
use Archive::Zip;
$obj = Archive::Zip->new(); # new instance
chdir 'Data/Temp/Files';
#files = <*>;
foreach $file (#files) {
$obj->addFile($file); # add files
}
$obj->writeToFileNamed("Images.zip");
The names and paths of zip archive members are completely independent of those of their real file counterparts. Although the two names are conventionally the same, AddFile allows you to specify a second parameter which is the name and path of the corresponding archive member where the file information should be stored
You can achieve the effect you're asking for my using basename from the File::Basename module to extract just the file name from the complete path
This program demonstrates. Note that it is essential to use strict and use warnings at the top of every Perl program you write
use strict;
use warnings;
use Archive::Zip;
use File::Basename 'basename';
my $zip = Archive::Zip->new;
for my $jpg ( glob 'Data/Temp/Files/*.jpg' ) {
$zip->addFile($jpg, basename($jpg));
}
$zip->writeToFileNamed('Data/Temp/Files/Images.zip');
Assuming I have the following array of file names I wish to zip
my #files = ("C:\Windows\Perl\test1.txt", "C:\Windows\Perl\test2.txt", "C:\Windows\Perl\test3.txt");
If I do
$obj = Archive::Zip->new();
foreach (#files)
{
$obj->addFile($_);
}
$obj->writeToFileNamed("zippedFolders.zip");
When I open zippedFolders.zip I see it contains subfolders, namely Windows and Perl, the latter which actually contains test1, test2, and test3. For some reason, the folders themselves are getting zipped up.
How can I make it so that only the files get zipped up and not have to click Windows then Perl folders to access the zipped files?
As you have see, if you use addFile to add a disk file to an archive, then Archive::Zip adds an archive member with the same path as file had originally. If you want it stored as something different then you can pass a second parameter that will be used as the name and path of the archive member created.
For your purposes I suggest you use the core File::Basename module to remove the path from the filename and pass just the basename as the second parameter.
The code below demonstrates.
Something else you need to be aware of is that you can't use single backslashes in Perl double quotes - they will be seen as escaping the following character and simply disappear. You can use pairs of backslashes in the string, use single quotes instead, use forward slashes instead (Perl will sort things out) or if there are no spaces in the filenames then you can use qw() as I have in this program.
use strict;
use warnings;
use Archive::Zip;
use File::Basename 'basename';
my #files = qw/
C:\Windows\Perl\test1.txt
C:\Windows\Perl\test2.txt
C:\Windows\Perl\test3.txt
/;
my $zip = Archive::Zip->new;
foreach my $file (#files) {
my $member = basename $file;
printf qq{Adding file "%s" as archive member "%s"\n}, $file, $member;
$zip->addFile($file, $member);
}
$zip->writeToFileNamed('zippedFolders.zip');
output
Adding file "C:\Windows\Perl\test1.txt" as archive member "test1.txt"
Adding file "C:\Windows\Perl\test2.txt" as archive member "test2.txt"
Adding file "C:\Windows\Perl\test3.txt" as archive member "test3.txt"
I have a directory or folder consisting of hundreds of files. They are named and arranged alphabatically. I want to move the files into directories or folders according to the first character of their name (i.e. files starting with a into one folder, files starting with r into another folder, etc).
Is there a way to do it without using CPAN modules?
Are the files all in that one folder, or are they in subfolders? If they are all in a single folder, you can use opendir to access the directory, and then readdir to read the file names and copy them elsewhere (using File::Copy module's move or copy function.
use strict;
use warnings;
use autodie;
use File::Copy; #Gives you access to the "move" command
use constant {
FROM_DIR => "the.directory.you.want.to.read",
TO_DIR => "the.directory.you want.to.move.the.files.to",
};
#Opens FROM_DIR, ao I can read from it
opendir my $dir, FROM_DIR;
# Loopa through the directory
while (my $file = readdir $dir) {
next if ($file eq "." or $file eq "..");
my $from = FROM_DIR . "/" . "$file";
move $from, TO_DIR;
}
This doesn't do exactly what you want, but it should give you the idea. Basically, I'm using opendir and readdir to read the files in the directory and I'm using move to move them to another directory.
I used the File::Copy module, but this is included in all Perl distributions, so it's not a CPAN module that must be installed.
Use glob(), or the built-in File::Find to build a list of files for each starting letter.
I want to locate the latest subdirectory on a network path and copy the entire contents of the latest subdirectory into another folder in the network path
We have lot of subfolders under the folder \\10.184.132.202\projectdump I need to sort the sub folders to get into latest folder and copy the entire contents into another folder on \\10.184.132.203\baseline
I am using the below mentioned script i am able to list the latest modified folder under the directory but I am unaware of copying the contents.
use File::stat;
use File::Copy qw(copy);
$dirname = '\\\\10.184.132.202\\projectdump\\Testing\\';
$destination = '\\\\10.184.132.203\\baseline\\Testing\\';
$timediff=0;
opendir DIR, "$dirname";
while (defined ($sub_dir = readdir(DIR)))
{
if($sub_dir ne "." && $sub_dir ne "..")
{
$diff = time()-stat("$dirname/$sub_dir")->mtime;
if($timediff == 0)
{
$timediff=$diff;
$newest=$sub_dir;
}
if($diff<$timediff)
{
$timediff=$diff;
$newest=$sub_dir;
}
}
}
print $newest,"\n";
open my $in, '<', $newest or die $!;
while (<$in>) {
copy *, $destination; --------> Here i want to copy the entire contents of the $newest to $destination.
}
Use File::Copy::Recursive. This is an optional module, but allows you to copy entire directory trees. Unfortunately, File::Copy::Recursive is not a standard Perl module, but you can install it via the cpan command.
If installing modules is a problem (sometimes it is), you can use the File::Find to go through the directory tree and copy files one at a time.
By the way, you can use forward slashes in Perl for Windows file names, so you don't have to double up on backslashes.
Why don't call a simple shell cmd to find the latest dir?
I think, this will be much simpler in shell...
my $newestdir=`ls -1rt $dirname|tail -n 1`;
in shell:
LATESTDIR=`ls -1rt $dirname|tail -n 1`
cp -r ${LATESTDIR}/* $destination/
Ups, I just realized that you might using Windows...
Get all dirs and their times into a hash then sort that hash reverse order to find the newest one
my ($newest) = sort {$hash{$b} cmp $hash{$a} keys %hash;
then
opendir NDIR, "$newest";
while ($dir=<NDIR>) {
next if $dir eq '.' or $dir eq '..';
copy $dir, $destination;
}
I want to uncompress zipped file say, files.zip, to a directory that is different from my working directory.
Say, my working directory is /home/user/address and I want to unzip files in /home/user/name.
I am trying to do it as follows
#!/usr/bin/perl
use strict;
use warnings;
my $files= "/home/user/name/files.zip"; #location of zip file
my $wd = "/home/user/address" #working directory
my $newdir= "/home/user/name"; #directory where files need to be extracted
my $dir = `cd $newdir`;
my #result = `unzip $files`;
But when run the above from my working directory, all the files get unzipped in working directory. How do I redirect the uncompressed files to $newdir?
unzip $files -d $newdir
Use Perl command
chdir $newdir;
and not the backticks
`cd $newdir`
which will just start a new shell, change the directory in that shell, and then exit.
Though for this example, the -d option to unzip is probably the simplest way to do what you want (as mentioned by ennuikiller), for other types of directory-changing, I like the File::chdir module, which allows you to localize directory changes, when combined with the perl "local" operator:
#!/usr/bin/perl
use strict;
use warnings;
use File::chdir;
my $files= "/home/user/name/files.zip"; #location of zip file
my $wd = "/home/user/address" #working directory
my $newdir= "/home/user/name"; #directory where files need to be extracted
# doesn't work, since cd is inside a subshell: my $dir = `cd $newdir`;
{
local $CWD = $newdir;
# Within this block, the current working directory is $newdir
my #result = `unzip $files`;
}
# here the current working directory is back to what it was before
You can also use the Archive::Zip module. Look specifically at the extractToFileNamed:
"extractToFileNamed( $fileName )
Extract me to a file with the given name. The file will be created with default modes. Directories will be created as needed. The $fileName argument should be a valid file name on your file system. Returns AZ_OK on success. "