如何根据MongoDB中的唯一ID获取多个属性？

Question

I'd like to retrieve say a list of only ids and genders as list<DBObject> from

{ "_id": 123 , "gender": m, "occupation": athlete },
{ "_id": 456 , "gender": f, "occupation": basketball },
...

Tried using the following but did not work. With query being initiated as a new BDObject gender : 1

dbcollection.distinct("_id", query)

Answer 1

db.getCollection('mobiledashboards').find({"_id":ObjectId("58d22f68acfecc105733d56a")})

Answer 2

"_id" is unique, so no need to call distinct function on it

If you want to get all the documents with gender and _id you can achieve it in following way

BasicDBObject query = new BasicDBObject();
BasicDBObject selectGenderOnly = new BasicDBObject();
selectGenderOnly.put("gender", 1);

DBCursor cursor = dbcollection.find(query, selectGenderOnly); // db.collection.find({}, {"gender": 1}) & _id field included by default in result set

 while (cursor.hasNext()) {
    BasicDBObject obj = (BasicDBObject) cursor.next();
    // use obj
    // ...
  }

It will provide the documents like below

{"_id":123 , "gender": m}
{"_id":456 , "gender": f}