Update: Oops, misread the question. Clearly not the answer.
Update2: This should do it though :-)
Something like this?
use warnings;
use strict;
my %count;
my $hash = {Grand => { Parent1 => {Child1 => { name => "me",
age => 99 },
Child2 => { name => "you",
age => 42 },
Child3 => { name => "she",
age => 1 }}}};
recurse_or_count($hash->{Grand},"Grand");
sub recurse_or_count {
my ($cur,$string)=@_;
for my $elem (keys %$cur) {
if (ref($cur->{$elem}) eq "HASH") {
$count{$string}++;
recurse_or_count ($cur->{$elem},$string."->$elem");
} else {
$count{$string}->{$elem}=$cur->{$elem};
}
}
}
for my $key (sort keys %count) {
if (ref($count{$key}) eq "HASH") {
for my $data (sort keys %{$count{$key}}) {
print "$key has $data $count{$key}->{$data}\n";
}
}else {
my $child = $count{$key} > 1 ? "children" : "child";
print "$key has $count{$key} $child\n";
}
}
__OUTPUT__
Grand has 1 child
Grand->Parent1 has 3 children
Grand->Parent1->Child1 has age 99
Grand->Parent1->Child1 has name me
Grand->Parent1->Child2 has age 42
Grand->Parent1->Child2 has name you
Grand->Parent1->Child3 has age 1
Grand->Parent1->Child3 has name she
Update3: and if you don't mind the child data being printed out before the child count, here's a more efficient version of that:
use warnings;
use strict;
my %count;
my $hash = {Grand => { Parent1 => {Child1 => { name => "me",
age => 99 },
Child2 => { name => "you",
age => 42 },
Child3 => { name => "she",
age => 1 }}}};
recurse_or_count($hash->{Grand},"Grand");
sub recurse_or_count {
my ($cur,$string)=@_;
for my $elem (sort keys %$cur) {
if (ref($cur->{$elem}) eq "HASH") {
$count{$string}++;
recurse_or_count ($cur->{$elem},$string."->$elem");
} else {
print "$string has $elem $cur->{$elem}\n";
}
}
}
for my $key (sort keys %count) {
my $child = $count{$key} > 1 ? "children" : "child";
print "$key has $count{$key} $child\n";
}
__OUTPUT__
Grand->Parent1->Child1 has age 99
Grand->Parent1->Child1 has name me
Grand->Parent1->Child2 has age 42
Grand->Parent1->Child2 has name you
Grand->Parent1->Child3 has age 1
Grand->Parent1->Child3 has name she
Grand has 1 child
Grand->Parent1 has 3 children
Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it. -- Brian W. Kernighan
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