You only need to have the variables evaluate to the keys. What you try doesn't give the compiler enough clues, it thinks the variable is to be the name of a method.

my $top = 'person';
my $foo = 'name';
print $ref->{$top}{$foo};
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After Compline,
Zaxo

Hi,

But i want to access the hash reference through a
variable as i don't know how deep it may go!

Like the following:



sub add_folder("Linux/Kernel")
{
    # i want to do the foll.
    # $bkref->{folder}{Linux}{folder}{Kernel}{folder} = {};
}
[download]


So, the RHS may go to any size
and i can't hard code it
Can u help me out?

--
arc_of_descent

As hint:

my $path = 'A/B/C';
my $hr = { 'A' => {'B'=>{'C'=>'Hello'}}};

my @path = split m</>,$path;
my $ref = $hr;
while(@path){
  $ref = $ref->{ shift @path }
}
print "$ref\n"
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--
http://fruiture.de

$ref = { "person" => {
                       "name" => "Bob"
                     }
};

# this works
print $ref->{person}{name};

# this does
$n = '{person}{name}';
eval 'print $ref->'.$n;
[download]

update: Just because you've been shown you have to use a knife, doesn't mean you also have to be told "Hey, don't cut your fingers off."

I did not forget, everybody knows its inherently dangerous.

____________________________________________________
** The Third rule of perl club is a statement of fact: pod is sexy.

Your answer is certainly right, but please don't forget to point out all the madness that lies that way when you post something like that.

Update: everyone knows things you can cut yourself with are dangerous. But not everyone knows which things you can cut yourself with, yet.

Makeshifts last the longest.