In your first code above, $wo_usr_proj{$proj} is a reference to an array, not an array. You need to dereference this value, as you did in your call to push: 
push @{$wo_usr_proj{$project}}, $user;

[download]
Currently the sort call is only sorting a list with one value (the address of the array stored in $wo_usr_proj{$proj}). The code below corrects this by dereferencing the hash element to retrieve the array: 
foreach $proj(sort(keys(%wo_usr_proj)))
{
   print RHANDLE "$proj:\n";$wo_usr_proj{$proj}

   # note dereference: @{ $array_ref_here }
   foreach $element(sort @{ $wo_usr_proj{$proj} })
   {
      print RHANDLE "$element\n";
   }
   print RHANDLE "\n\n";
}

[download]
Please take a look at Data::Dumper, as it is invaluable in illustrating the structure of the data in your code. You simply pass the Dumper function a reference to the data structure you wish to see expanded: 
use Data::Dumper;
print Dumper \%wo_usr_proj;

__END__
$VAR1 = {
          'project2' => [
                          'WO',
                          'WO2
                        ],
          'project1' => [
                          'WO'
                        ]
        };

[download]
As for your second code snippet, the code is not dereferencing $element, but rather, is creating another reference to $element.

If $scalar is a reference to a scalar value, dereference using $$scalar: 
my $foo= "bar";
my $ref= \$foo;

print $$ref;

__END__
bar

[download]
If $arr is a reference to an array, dereference using @$arr: 
my @array= qw(foo bar baz);
my $ref= \@array;

print join ':' => @$ref;

__END__
foo:bar:baz

[download]
If $href is a reference to a hash, dereference using %href: 
my %h= ( foo => 'bar' );
my $ref= \%h;

while( my( $k, $v )= each %$ref ) {
  print "$k -> $v"
}

__END__
foo -> bar

[download]
Please see the pod for perlreftut, perldsc, and perllol for more details.

--sacked

In reply to Re: Re: Re: dealing with arrays as hash values by sacked
in thread dealing with arrays as hash values by goonarific

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