Notwithstanding the fine answer you've already received, if you still want to do it with a fetchall, you can't build the hash with map (okay, you probably can, but not in a conventional way) because you're not building a straightforward hash, where each input element corresponds to a hash member. Instead, you're building a compound hash where each input element gets pushed onto a hash member.

This example code does the transformation you want.

#!perl
use Data::Dumper;

$rows = [[qw(1234 foo fooval)],[qw(1235 bar barval)],[qw(1234 quux quu
+xval)]];
my %hash;
for (@$rows) {
  push @{$hash{$_->[0]}}
  , { option      => $_->[1]
    , option_data => $_->[2]
    }
  ;
}

print Dumper(\%hash);
[download]

Update: which, of course, looks very similar to dragonchild's solution, because it's doing basically the same thing.