SomeNoob has asked for the wisdom of the Perl Monks concerning the following question:

Hello. This is my first question so my apologies in advance if I break any taboos. I did read the formatting tips and how to ask a question articles.

From a high level I am trying to write an XML parser. I'm using XML::Simple to get the XML into a hash and am then navigating that hash to pull out the data I need. In the hash that XML::Simple gives me, there is an array of hashes, and while looping through those I have a problem with the structure below:

'user' => {
                                                                      
+           'firstname' => 'Joe',
                                                                      
+           'lastname' => 'Shmoe',
               },
'user' =>  'Finance Department'

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What I want to do is print out either the user's full name OR the department if there is no substructure under the user reference. However, I get a "Can't use string ("Vox Procurement Managers") as a HASH ref while "strict refs" in use at Code.pl line 138." error when the code gets to the Finance Department bit. I've tried checking if(exists $xml->$usernumber->{"user"}->{"firstname"} before trying to print the full name but that still throws the error. Further googling and searching here has not helped. Basically I need a way to check if that reference in the hash points to a string ("Finance Department") or to more hashes down below. An if statement based on if( ref ($xml->$usernumber->{"user"}}) eq "HASH") always returned true and did not work either.

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Re: Parsing hashes with variable structure
by jethro (Monsignor) on Nov 09, 2012 at 01:55 UTC

    Is your problem solved with the addition of {} around $usernumber or was that a copy mistake? If the problem still exists, read on:

    #!/usr/bin/env perl
use warnings;
use strict;
use 5.10.0;

my $h= { 'user' => { 'firstname' => 'Joe','lastname' => 'Shmoe'}};

my $h2= { 'user' => 'Finance Department'};


say "yes" if (ref($h->{'user'}) eq "HASH");

say "no" if (not ref($h2->{'user'}) eq "HASH");

#OUTPUT:
yes
no

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    As you can see ref() works. So probably you feed the wrong data to ref(). You could use Data::Dumper to print out your $xml to see if it fits your expectation. And print out $xml->{$usernumber}->{"user"} just before your "if"-clause to check what it contains at that crucial moment. If ref() returns "HASH" then you can be really sure that it is a hash that is in there and not the string "Finance Department"

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      Thanks so much! ref() did work; I must have made some kind of mistake when I tried it before. Again, thanks a ton for helping me out.

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Re: Parsing hashes with variable structure
by frozenwithjoy (Priest) on Nov 09, 2012 at 00:27 UTC
    What do line 138 and relevant surrounding lines look like?
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      Line 138 itself is the if(exists...) line and it's in the middle of a loop going through the array of references to different user hashes.

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        Perhaps you meant the following? (I added {} around $usernumber) 
        if( exists $xml->{$usernumber}->{"user"}->{"firstname"} ) {...}

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        EDIT: You may also need to extend it as such: (but it is difficult to tell for sure w/o seeing more context)

        if( ref eq 'HASH' and exists $xml->{$usernumber}->{"user"}->{"firstnam
+e"} ) {...}

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