I\'m using mongo 2.2.3 and the java driver. My dilemma, I have to $push a field and value into an array, but I cant seem to figure out how to do this. A sample of my data:<
DBObject listItem = new BasicDBObject("scores", new BasicDBObject("type","quiz").append("score",99));
DBObject updateQuery = new BasicDBObject("$push", listItem);
myCol.update(findQuery, updateQuery);
MongoDB Java driver can simplify this. Use $each instead of $push.
$each mongodb reference document
Java sample -
BasicDBObject addressSpec = new BasicDBObject();
addressSpec.put("id", new ObjectId().toString());
addressSpec.put("name", "one");
BasicDBObject addressSpec2 = new BasicDBObject();
addressSpec2.put("id", new ObjectId().toString());
addressSpec2.put("name", "two");
List<BasicDBObject> list = new ArrayList<>();
list.add(addressSpec); list.add(addressSpec2);
UpdateResult updateOne = individualCollection.updateOne(Filters.eq("_id", "5b7c6b612612242a6d34ebb6"),
Updates.pushEach("subCategories", list));
Using Jongo, you can do as in the shell:
db.collection.update({_id:1},{$push:{scores:{type:"quiz", score:99}}})
Becomes in Java:
collection.update("{_id:1}").with("{$push:{scores:{type:#, score:#}}}", "quiz", 99);
No fancy DBObject needed ;-)
Since mongodb-driver 3.1.
there is a builder class com.mongodb.client.model.Updates
with appropriate methods for each update case. In this case this would be:
Document score = new Document().append("type", "quiz")
.append("score",99);
collection.updateOne(eq("_id", "1"),Updates.addToSet("scores", score));
If you're more comforable with the query format of the shell, you may find it's easier to use JSON.parse
to contstruct your DBObject
for the $push
:
import com.mongodb.util.JSON;
String json = "{$push:{scores:{type:'quiz', score:99}}}";
DBObject push = (DBObject) JSON.parse(json);