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

The Enlightened,

perl v5.8.8, use strict, use diagnostics

I hava an array of hashes that match to database fields 

$customer->[0]{_name}
$customer->[0]{_address1}
$customer->[0]{_address2}

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I would like to substitute the hash name with 
my $i = 1;
$varname = "_address" . $i;
print "Customer Address", $customer->[0]{$varname};

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for looping purposes, but it doesn't work. 
print "Customer Address", $customer->[0]{${varname}};

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does not work. I get the an uninitialized value error (for both tries).

I could put the hashes into address array and then loop through those but laziness makes me think perl can do it...

Thank you,

Replies are listed 'Best First'.
Re: Hash name variable substitution
by Errto (Vicar) on May 21, 2007 at 20:13 UTC
    The following works fine for me: 
    $customer->[0]{_address1} = 'hello';
my $i = 1;
my $varname = '_address' . $i;
print $customer->[0]{$varname};

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    prints "hello". Perhaps your issue lies elsewhere. If you can come up with a small runnable program that illustrates this error that may be more helpful.
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[d/l]

      Thanks Errto,

      Sorry - the database field has no entry hence the uninitialized value. Choosing a populated field works fine. I will just take my slice of humbleberry pie for wasting the Enlightened's time...

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Re: Hash name variable substitution
by thezip (Vicar) on May 21, 2007 at 20:28 UTC

    Hey peterb, are you looking for something like this?

    #!/perl/bin/perl -w
use strict;
use Data::Dumper;

my $data = [
  { _name     => "John Smith",
    _address1 => "123 Foo St.",
    _address2 => "Apt 5",
  },
  { _name     => "Brigitte Nelson",
    _address1 => "150 Park Pl.",
    _address2 => "Suite 223",
  },        
];


my @cols = qw / _name  _address1  _address2  /;

for my $customer (@$data) {
  for my $col (@cols) {
    printf "%s => %s\n", $col, $customer->{$col} || "<empty>";
  }
}

    [download]

    Updated: Added <empty> string in case of emptiness...


    Where do you want *them* to go today?
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[d/l]
Re: Hash name variable substitution
by shmem (Chancellor) on May 21, 2007 at 20:33 UTC
    I would like to substitute the hash name with

    You don't have a hash name - which would mean you had named hashes - but hash keys in your array of anonymous hashes. If you change the key and reference that changed key, there's no value behind it - the value you are trying to acces is bound to the old (unchanged) key. To change a key, you have to copy over the value the old key is pointing at into the slot pointed at by the new key: 

    my $i = 1;
$varname = "_address" . $i;
$customer->[0]{$varname} = $customer->[0]{'_address'};
print "Customer Address", $customer->[0]{$varname};

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    But... 

    $customer->[0]{_address1}
$customer->[0]{_address2}

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    whenever there are keys with a number attached, that cries for a revision of the data structure: store an anonymous array as the value to that hash. Then you have 

    $customer->[0]{_address}[0]
$customer->[0]{_address}[1]

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    and looping becomes much easier: 

    my $arrayref = $customer->[0]{_address};
for my $address (@$arrayref) {
    # do whatever with $address
    ...
}

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    But if you really have those column names (_address1,_addresss2,...,_adressn) in your tables, then the database design is flawed. You might want to normalize the addresses out to a another table.

    --shmem 

    _($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}
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[d/l]
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