Renaming a Zip file with a user input variable

mayank646 has asked for the wisdom of the Perl Monks concerning the following question:

Hi all, I am trying to rename a zip file with the user input. I have tried but when i am giving the variable its making a file with the vairiable name only like $abc not with the value stored in the variable.

#!/bin/perl

use strict;
use warnings ; 
use Archive::Zip; 

my $zip_file = 'file.zip' ;

my $zip = Archive::Zip-> new (); 
$zip ->addTree( 'c:/zip'); 
$zip ->writeToFileNamed( $zip_file ); 
if (-e $zip_file) {
    print "Archive created successfully!";
}    else{
    print "Error in archive creation!";
}
[download]

Instead of $zip_file being hardcoded i want to take user input for the zip file name. I am using the following code

#! usr/bin/perl

use Archive::Zip;

# Getting Details 
print "Enter the Folder Name \n";
$rn = <>;
print "The details are as follows: \n";
print "Release Name is *** $rn *** \n";
print "Business Unit is *** ABC *** \n";
print "Environment is *** ABC *** \n";

# Taking Backup 

my $zip_file = 'file.zip';

my $zip = Archive::Zip-> new (); 
$zip ->addTree( 'c:/zip'); 

$zip ->writeToFileNamed($zip_file); 
if (-e $zip_file) {
    print "Archive created successfully!";
}    else{
    print "Error in archive creation!";
}
[download]

here i am taking the input from the user and storing it in a variable $rn now i want to make the zip file with the name stored in the variable $rn.

Comment on Renaming a Zip file with a user input variable Select or Download Code

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Re: Renaming a Zip file with a user input variable by toolic (Bishop) on Aug 06, 2012 at 18:30 UTC
shift, then call your script with the file name as an argument. `my $zip_file = shift;` [download]	[reply] [d/l]
Re: Renaming a Zip file with a user input variable by aitap (Curate) on Aug 06, 2012 at 19:46 UTC
See @ARGV, Getopt::Std, Getopt::Long. For example, `my $file = $ARGV[0] \|\| die "Usage: $0 <file>\n";` Sorry if my advice was wrong.	[reply] [d/l]
Re: Renaming a Zip file with a user input variable by rpnoble419 (Pilgrim) on Aug 06, 2012 at 21:33 UTC
Where are you getting the the user input from when you go into production this code? You do not want to use any web form data directly as this may open your code up to a security issue.	[reply]
Re^2: Renaming a Zip file with a user input variable by mayank646 (Initiate) on Aug 07, 2012 at 12:17 UTC
Hi I am using the following code `#! usr/bin/perl use Archive::Zip; # Getting Details print "Enter the Folder Name \n"; $rn = <>; print "The details are as follows: \n"; print "Release Name is * $rn * \n"; print "Business Unit is * ABC * \n"; print "Environment is * ABC * \n"; # Taking Backup my $zip_file = 'file.zip'; my $zip = Archive::Zip-> new (); $zip ->addTree( 'c:/zip'); $zip ->writeToFileNamed($zip_file); if (-e $zip_file) { print "Archive created successfully!"; } else{ print "Error in archive creation!"; }` [download] here i am taking the input from the user and storing it in a variable $rn now i want to make the zip file with the name stored in the variable $rn.	[reply] [d/l]