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

I'm currently doing something like: 
$counts{$n{'Rep'}}{$n{'Item'}} += $n{'Qty'};

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This is fine if I always sort the data this way, but what if I decide I want to use a 3-level hash without hard-coding the data assignment? How do I efficiently take an array of keys: 
@keys = ('Rep','Item','Whatever');

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And translate it into: 
$counts{$n{'Rep'}}{$n{'Item'}}{$n{'Whatever'}} += $n{'Qty'};

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It doesn't have to be neat, or the fastest code on the planet, but efficiency is somewhat of a concern. I suppose I could use eval (since this is an admin script and placed safely out of reach of hackers), but I'm interested in seeing if there are other ways.

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Re: Assigning data to a nested hash with a variable number of levels?
by tlm (Prior) on Apr 19, 2005 at 03:49 UTC

    This node (and its follow-ups) may be of interest. And this one.

    the lowliest monk

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Re: Assigning data to a nested hash with a variable number of levels?
by TedPride (Priest) on Apr 19, 2005 at 04:36 UTC
    The module solution would probably work fine (assuming my client can figure out how to install the proper module), but the other post seems a bit confusing. It doesn't matter though - I finally figured out how to do it properly with references (or is it pointers?): 
    use strict;
use warnings;

my %hash;
assign(\%hash, 'aa', 'bb', 'cc', 'value');
print $hash{aa}{bb}{cc};

sub assign {
    my $p = shift;
    my $val = pop;
    my $last = pop;
    for (@_) {
        $p->{$_} = {} if !$p->{$_};
        $p = $p->{$_};
    }
    $p->{$last} = $val;
}

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    It came to me while I was lying in bed reading a book. Odd how that works.
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[d/l]

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