my %hash2 = map { $_ => $hash1{$_} }
            grep { defined $hash1{$_} } @ary;
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It came to me about a minute after you'd posted, after about an hour of faffing, thanks anyway!!!!
grep is the answer!

my %hash2 = map {defined($hash1{$_}) ? ($_ => $hash1{$_}) : ()} @ary;
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For a quick, but possibly wrong solution, you could use a hash slice:

my %newhash;
@newhash{@array} = @oldhash{@array};
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A pitfall here is that if there's a value in the array that doesn't correspond to a key in the original hash, you'll autovivify an entry.

If you don't have such control over your data, try:

my %new = map { $_=> $old{$_} } grep { defined $old{$_} } @array;
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The grep gets all the entries in the hash that (a) have a key corresponding to an entry in the array and (b) have defined values. The map turns that list into a hash.

Change that defined to exists if you don't mind getting back keys associated with undefined values.

HTH!

perl -e 'print "How sweet does a rose smell? "; chomp ($n = <STDIN>); 
+$rose = "smells sweet to degree $n"; *other_name = *rose; print "$oth
+er_name\n"'
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Autovivification does not occur here; it only happens when the non-existent thing in question is an lvalue:

%hash = qw( a 1 b 2 c 3 );
@array = (1,2,4,8);

$x = $hash{z};    # no auto-viv
$x = $array[10];  # no auto-viv

1 for $hash{y};   # auto-viv
1 for $array[7];  # auto-viv
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Bonus: why are those bottom two treating the value in question as an lvalue?

_____________________________________________________
Jeff japhy Pinyan: Perl, regex, and perl hacker.
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

looks like because they're being evaluated in list context - but I don't even know what vivification is, let alone autovivification

I'm off to look it up.

update: hell, I can't even spell it!

update2: found a useful article here - turns out I do most of this already - I just didn't know what it was called!

update3: for supersearch: "what is autovivification?" and for muppets like me in the supersearch "what is autovivication?"

"Argument is futile - you will be ignorralated!"

my %hash2;
for (@ary) {
    $hash2{$_} = $hash1{$_} if exists $hash1{$_};
}
undef %hash1;
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UPDATED For the reason that the array is coming in from user input (Getopt::Long), so I don't want to add unknown options to my final hash.
I want to pare down the entries in the hash to the mimimum amount of keys corresponding to only known parameters that i have been given.
cheers though, Sorry the first draft of this reply was gibberish

my %opts;
my %locs = map { $_ => 1 } ('do', 'ny', 'kw');
my %sect = ('uid' => 1, 'email' => 'a', 'username' => 'a');

my $USAGE = qq(

USAGE: perl launch.pl (-l=<location>)+ (-d=<data>)+

Where -l=<location>  = [do|kw|ny]           = one or more times
      -d=<data_type> = [uid|email|username] = one or more times
);

die $USAGE unless (scalar(@ARGV) >= 2);

GetOptions(\%opts, "l=s@", "d=s@");

die $USAGE unless (defined $opts{'l'});
die $USAGE unless (defined $opts{'d'});

%locs = map { $_ => $locs{$_}} grep defined $locs{$_},  @{ $opts{'l'} 
+};   # strips out keys from hash if
%sect = map { $_ => $sect{$_}} grep defined $sect{$_},  @{ $opts{'d'} 
+};   # not requested by user

die $USAGE unless scalar keys %locs > 0;
die $USAGE unless scalar keys %sect > 0;
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Umm... you won't be. The only keys in %hash1 that are ever checked are values that exist in @ary. I'm assuming that @ary is the list of allowable parameters. That would mean that any values in %hash1 that didn't exist in @ary would never make it into %hash2. And, likewise, any values in @ary that don't exist in %hash1 don't make it into %hash2.