I was wondering if it was possible to use the "in" operator as you can from the mongo shell, using the perl MongoDB::Collection module. I have tried a number of things, but haven't quite got the result I am expecting. I've check the docs and other posts on stackoverflow but can't seem to find anything specifically about this, unless I am overlooking something.
http://docs.mongodb.org/manual/reference/operator/query/in/
The count query I am running via the mongo shell is
mongo:PRIMARY> db.getCollection("Results").count( { TestClass : "TestClass", TestMethod : { $in: ["method1" , "method2", "method3"] } })
181605
I have tried this a few different ways passing the list as an array or hash-refs or pre-building a string...
my $count = $mongo->{collection}->count({
'TimeStamp' => { '$gt' => $ft, '$lt' => $tt },
'TestClass' => $TestClass,
'TestMethod' => { '$in' => [$whitelist->methods] },
'Result' => $result
});
Where Dumping $whitelist->methods
is
$VAR1 = {
'method1' => 1,
'method2' => 1,
'method3' => 1
};
I've looked high and low for an answer, does anyone know if the driver is currently capable of using the $in operator like this? Looping through the returned methods from a previous query and adding up the results will require more code.
The only other stack overflow post I have seen about the $in operator was this $in mongoDB operator with _id in perl recommending using http://api.mongodb.org/perl/current/MongoDB/OID.html but don't think that is relevant in my example as looks more to do with ID's.
Any help or discussion would be greatly appreciated.
The problem is that $in
clause expects its value to be an array reference, but you supply a hashref (as Dumper's output shows) into it. The easiest way to turn the latter into the former is to apply keys
function:
# ...
'TestMethod' => { '$in' => [keys %{$whitelist->methods}] }
... or just [keys $whitelist->methods]
, if you're using Perl 5.14+, as ...
starting with Perl 5.14,
keys
can take a scalar EXPR, which must contain a reference to an unblessed hash or array
.