in reply to Hash name variable substitution

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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Re^2: Hash name variable substitution
by peterb (Novice) on May 21, 2007 at 20:24 UTC

    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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