Re: Renaming a Zip file with a user input variable

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.

Comment on Re: Renaming a Zip file with a user input variable

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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]

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Re^2: Renaming a Zip file with a user input variable
by mayank646 (Initiate) on Aug 07, 2012 at 12:17 UTC

#! 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!";
}
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