in reply to hash with a hash

You probably want 

$tests{2} = \%blah2;

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(i.e., you need to assign a hashref when you're in scalar context...  Otherwise, you get the number of used/allocated buckets of the hash — which is why you see the error 'Can't use string ("2/8") as a HASH ref while "strict refs" in use...')

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Re^2: hash with a hash
by perlfan (Parson) on Jun 04, 2007 at 19:34 UTC
    Yeah, 
    $tests{2} = \%blah2;

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    is the same as 
    $tests{2} = { 
              condition => "AND",
              terms => [ "foo", "bar" ],
            };

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    One thing that you should note is that when explicitly defining hashes and arrays, one can use "()"'s. 
    my @array = ( 1, 2, 3 );
my %hash = ( A => 1, B => 2, C => 3, );

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    However, anonymous arrays are defined with "[]"'s 
    my $array_ref = [ 1, 2, 3 ];
my @array = @{$array_ref}; # becomes an explicit array

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    and anonymous hashes are defined with "{}"'s. 
    my $hash_ref = { A => 1, B => 2, C => 3, };
my %hash = %{$hash_ref}; # becomes an explicit hash

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