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09.11.2007 at 10:34AM PDT, ID: 22821138
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Perl Copy directory using variables.

Tags: perl, copy, directory
Need some help...
I've created a variable, $StdFolder, which points to a directory that I want to use as the standard for all new projects. $StdFolder = 't:/projects/standard/';

I want to take that directory an copy the folder structure of that directory into a folder that was created in the step before.

I'm looking for something like  the follwoing: xcopy "$StdFolder" / "$c1002 $Proj", but I don't know how to make it work.

Here's my code so far.

use CGI;
use strict;

my $obj = new CGI;
my $Proj = $obj->param ( 'Project' );
my $StdFolder = 't:/projects/standard/';
my $c1002 = 't:/projects/concepts/1002xxx/';
my $c1003 = 't:/projects/concepts/1003xxx/';
my $c1004 = 't:/projects/concepts/1004xxx/';

print "Content-type: text/html\n\n";
if ($Proj =~ /1002\d\d\d/)
{
mkdir ( $c1002.$Proj, 0777 );
print "Your1002xxx promotional project, $Proj, has been created.\n"
}

elsif ($Proj =~ /1003\d\d\d/)

{
mkdir ( $c1003.$Proj, 0777 );
print "Your 1003xxx promotional project, $Proj, has been created.\n"
}

elsif ($Proj =~ /1004\d\d\d/)

{
mkdir ( $c1004.$Proj, 0777 );
print "Your 1004xxx promotional project, $Proj, has been created.\n"
}
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Question Stats
Zone: Programming
Question Asked By: Scudboy
Solution Provided By: ozo
Participating Experts: 4
Solution Grade: B
Views: 103
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09.11.2007 at 11:05AM PDT, ID: 19870752

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09.11.2007 at 11:09AM PDT, ID: 19870794

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09.11.2007 at 11:29AM PDT, ID: 19870976

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09.11.2007 at 11:36AM PDT, ID: 19871052

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09.11.2007 at 11:39AM PDT, ID: 19871077

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09.11.2007 at 12:29PM PDT, ID: 19871487

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09.11.2007 at 02:30PM PDT, ID: 19872403

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09.11.2007 at 06:32PM PDT, ID: 19873321

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09.12.2007 at 07:05AM PDT, ID: 19876313

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09.12.2007 at 08:31AM PDT, ID: 19877171

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