in reply to dealing with arrays as hash values

You're not that far off, I think your code would work if you replaced line 8 with $wo_usr_proj{$project} = \@usr_list; - hash values must be scalars, so a reference is needed.

Your code can be cleaned up a bit. You can read data directly from the file, for (<INFH>) { Loop variables are better as lexicals, and split uses $_ as the default second argument,     my ($user, undef, $project) = split /\|/; The remainder can be condensed to a conditional push onto the dereferenced array in the hash, 

    push @{$wo_usr_proj{$project}}, $user
        if $uo_hash{lc($user)} =~ /WO/;
}

[download]

After Compline,
Zaxo

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Re: Re: dealing with arrays as hash values
by goonarific (Initiate) on May 17, 2004 at 21:01 UTC
    Thanks, your comments for cleaning up my code helped a lot. I switched over that entire three lines , 6-8, to 
    push @{$wo_usr_proj{$project}}, $user;

    [download]
    I had some other stuff in that if statement that needed to be done too. but now when I print out the values from %wo_usr_proj, I only get array memory addresses: 
    foreach $proj(sort(keys(%wo_usr_proj)))
{
   print RHANDLE "$proj:\n";
   foreach $element(sort($wo_usr_proj{$proj}))
   {
      print RHANDLE "$element\n";
   }
   print RHANDLE "\n\n";
}

    [download]
    That gave me the array memory adresses as output into my file I then tried: 
    foreach $proj(sort(keys(%wo_usr_proj)))
{
   print RHANDLE "$proj:\n";
   foreach $element(sort($wo_usr_proj{$proj}))
   {
      print RHANDLE \$element;
      print RHANDLE "\n";
   }
   print RHANDLE "\n\n";
}

    [download]
    thinking that by derefencing $element it would give me the value, but instead it gave me scaler memory adresses. I don't think that the user names are actually getting put into the hash. Is there some initialization that needs to occur to "tell" the hash that its value is an array? I think if I get past this last part I'll be done. Thank you in advance for any help or comments on my code.
[reply]
[d/l]
[select]
      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
[reply]
[d/l]
[select]

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