Solved

Parse/traverse nested HASH

Posted on 2006-06-23
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711 Views
Last Modified: 2008-03-10
I have following problem, i want one max two subs that call it self as long the HASH tree isn't ended, i have written following code:

#!/usr/bin/perl
#use strict;
use Scalar::Util qw(reftype);

my %hello;
$hello{'a1'}->{'a2'}->{'attr1'} = "abc1";
$hello{'a1'}->{'a2'}->{'attr2'} = "abc2";
$hello{'b1'}->{'b2'}->{'attr1'} = "bar1";
$hello{'b1'}->{'b2'}->{'attr2'} = "bar2";
$hello{'b1'}->{'b2'}->{'attr3'} = "bar3";

my $nk;
my $nnk;
my %nh;
my $nh;
&testen($nk);
my $counter = 0;
sub testen{
    $counter++;
    print "Count:$counter\n";
    if(!$nk eq ''){
      %nh = %{$nh{$nk}};
    }else{
      %nh = %hello;
    }
    foreach(keys %nh)
    {
      my $kk;
      $kk = $_;
      if(reftype($nh{$_}) eq 'HASH'){
          for(keys %{$nh{$_}})
          {
            print "KK:$kk S:$_\n";
            $nk = $kk;
            testen($counter, $nk, %nh);
          }
      }else{
          print "Value: $nh{$_}\n";
      }
    }
}
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Comment
Question by:pcl99
7 Comments
 
LVL 10

Expert Comment

by:xanius
ID: 16968462
pcl99,

it would be nice if you could tell us what your code is supposed to do/producde, then it is nuch easier to make a suggestion.

Cheers
Xanius
0
 

Author Comment

by:pcl99
ID: 16969578
I want to return the keys, but the HASH example can be different from time to time because it's gonna be input data from another source.

it could also look like this:
$hello{'a1'}->{'a2'}->{'attr1'}->{'facs'} = "abc1";
$hello{'a1'}->{'a2'}->{'attr2'}->{'tt'}->{'yyy'} = "abc2";
$hello{'b1'}->{'b2'}->{'attr1'} = "bar1";
$hello{'b1'}->{'b2'}->{'attr2'} = "bar2";
$hello{'b1'}->{'b2'}->{'attr3'} = "bar3";
$hello{'c1'} = "Test";

I need the values from it...
to design a new HASH where other values will be added, but it's here i have the problem.

Cheers
pcl99
0
 
LVL 8

Expert Comment

by:Perl_Diver
ID: 16970905
you want  the names of the keys (a1, b1 etc) or the values of the hash keys (abc1, abc2, etc)?
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Author Comment

by:pcl99
ID: 16971746
I got the solution myself, see here:

#!/usr/bin/perl
use strict;
use Scalar::Util qw(reftype);

my %hello;
$hello{'a1'}->{'a2'}->{'attr1'} = "abc1";
$hello{'a1'}->{'a2'}->{'attr2'} = "abc2";
$hello{'b1'}->{'b2'}->{'attr1'} = "bar1";
$hello{'b1'}->{'b2'}->{'attr2'} = "bar2";
$hello{'b1'}->{'b2'}->{'attr3'} = "bar3";

testen(\%hello);

sub testen{
    my $hRef = shift || return 0;
    my $level = shift || 0;
    $level++;
    return 0 if $level > 50;
    if (reftype($hRef) eq 'HASH') {
      foreach my $k (keys %{$hRef}) {
          print "Key:$k\n";
          testen($hRef->{$k}, $level);
      }
    }else{
      print "Value:$hRef\n";
    }
}
0
 
LVL 10

Expert Comment

by:xanius
ID: 17316615
Fine with me!
Xanius
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Accepted Solution

by:
CetusMOD earned 0 total points
ID: 17347263
PAQed with points refunded (250)

CetusMOD
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