Re^3: Compare two Hashes of Hashes

[kennethk]

Thank you for posting that code. Note that with the block

  for my $key_hash1 (sort {lc($a) cmp lc($b) } keys %{$hash1} ) {
    if( ! exists $hash2->{$key_hash1} ) {
      print "Hash1 contains unit '$key_hash1' but hash2 has no such un
+it\n";
      next;
    }

you will check if %hash2 is missing keys from %hash1, but you lack code to do the reverse check. So you probably want to add something like:

  for my $key_hash2 (sort {lc($a) cmp lc($b) } keys %{$hash2} ) {
    if( ! exists $hash1->{$key_hash2} ) {
      print "Hash1 contains unit '$key_hash2' but hash1 has no such un
+it\n";
    }
  }

For adding another layer of checks, you need to replace your value check

    if( $hash1->{$key_hash1} ne $hash2->{$key_hash1} ) {
      print "Both hashes have '$key_hash1' as a unit ",
            "but hash1's value is '$hash1->{$key_hash1}' ",
            "and hash2's value is '$hash2->{$key_hash1}'\n";
    }

with another loop over the keys of the hashes, very much like you've done already.

  for my $key_hash1 (sort {lc($a) cmp lc($b) } keys %{$hash1} ) {
    if( ! exists $hash2->{$key_hash1} ) {
      print "Hash1 contains unit '$key_hash1' but hash2 has no such un
+it\n";
      next;
    }
    for my $key_hash1_tier2 (sort {lc($a) cmp lc($b) } keys %{$hash1->
+{$key_hash1}} ) {
      if( ! exists $hash2->{$key_hash1}{$key_hash1_tier2} ) {
        print "Hash1 $key_hash1 contains unit '$key_hash1_tier2' but h
+ash2 has no such unit\n";
        next;
      }
      if( $hash1->{$key_hash1}{$key_hash1_tier2} ne $hash2->{$key_hash
+1}{$key_hash1_tier2} ) {
        print "Both hashes have '$key_hash1'-'$key_hash1_tier2' as a u
+nit ",
              "but hash1's value is '$hash1->{$key_hash1}{$key_hash1_t
+ier2}' ",
              "and hash2's value is '$hash2->{$key_hash1}{$key_hash1_t
+ier2}'\n";
      }
    }
  }

subject to the same criticism I made before about not checking which elements are in %hash2 but not %hash1. You can see this clearly if you reverse the order of the arguments to comp.