$report_data{$domain}{$provider} is a hash reference already. This is shown by your data dumper (which shows that it contains hash key/values).
You are then taking the reference to that hash reference by passing it as \$report_data{$domain}{$provider} which makes it a reference to a reference.
Hence, you need to either not pass it that way or dereference it twice.
$,=" : ";$\="\n"; print $_, $$rpt_ref->{$_} for keys %{$$rpt_ref};
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It's sad that a family can be torn apart by such a such a simple thing as a pack of wild dogs
| [reply]
[d/l] |
Thanks, I see that now. Although I don't understand how it is a reference at that point. Prior to that point it is declared and I've simply been populating it via:
my %report_data;
...
$report_data{$domain}{$provider}{ok}++;
| [reply]
[d/l] |
You don't understand how $report_data{$domain}{$provider} is a reference?
If not, if you were to define things all at once it would be like this:
my %report_data = (
$domain => {
$provider => {
'error' => '0',
... # other stuff here
}
}
}
The reason it's a reference is because you have a scalar value (hash value $report_data{$domain}) pointing to a hash reference.
From perldoc -f perldata:
...
All data in Perl is a scalar, an array of scalars, or a hash of scalars
...
Although a scalar may not directly hold multiple values, it may contain a reference to an array or hash which in turn contains multiple values.
...
A hash isn't a scalar, it's a hash, but a reference is a scalar, so it can hold that in the value of the hash where the key is $provider.
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It's sad that a family can be torn apart by such a such a simple thing as a pack of wild dogs
| [reply]
[d/l] |
