Re: In a hash with hashes

You're overwriting the hash value with each new patient. Try:

$practice = {}; 

foreach $patient(@{$self->{patient_details}})
{
     push @{ $practice->{ $patient->{suregery_id} } }, $patient;
}
[download]

It creates an array of patients for each id.

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.

Comment on Re: In a hash with hashes Download Code

Replies are listed 'Best First'.
Re^2: In a hash with hashes by AnomalousMonk (Archbishop) on Jan 10, 2010 at 21:57 UTC
Since BrowserUk has already provided the answer, this reply is very much BTW, but FWIW... In the OPed code fragment, the statement `$practice->{$patient->{suregery_id}} = \$patient;` repeatedly assigns (i.e., overwrites) a reference to `$patient` to the hash key `$patient->{suregery_id}` of the hash referenced by `$practice`. The OPed code iterates over an array of hash references, so `$patient` is already a hash reference. The statement above repeatedly assigns the LHS hash element a reference to a reference to a hash, which is probably even less what weezer_316 wanted! (Again, not that it matters much at this point.)	[reply] [d/l] [select]
Re^2: In a hash with hashes by weezer_316 (Novice) on Jan 10, 2010 at 13:56 UTC
The code is at work, I'm at home, but I think your a genius! Thats utterly obvious and ive totally missed it. Thanks alot!	[reply]