How to get selected object only from an array - spring

Question

I have a collection with documents of the following structure:
{
"category": "movies",
"movies": [
{
"name": "HarryPotter",
"language": "english"
},
{
"name": "Fana",
"language": "hindi"
}
]
}
I want to query with movie name="fana" and the response sholud be
{
"category": "movies",
"movies": [
{
"name": "HarryPotter",
"language": "english"
}
]
}
How do I get the above using spring mongoTemplate?

Answer 1

You can try something like this.
Non-Aggregation based approach:
public MovieCollection getMoviesByName() {
BasicDBObject fields = new BasicDBObject("category", 1).append("movies", new BasicDBObject("$elemMatch", new BasicDBObject("name", "Fana").append("size", new BasicDBObject("$lt", 3))));
BasicQuery query = new BasicQuery(new BasicDBObject(), fields);
MovieCollection groupResults = mongoTemplate.findOne(query, MovieCollection.class);
return groupResults;
}
Aggregation based approach:
import static org.springframework.data.mongodb.core.aggregation.Aggregation.*;
import static org.springframework.data.mongodb.core.query.Criteria.where;
public List<BasicDBObject> getMoviesByName() {
Aggregation aggregation = newAggregation(unwind("movies"), match(where("movies.name").is("Fana").and("movies.size").lt(1)),
project(fields().and("category", "$category").and("movies", "$movies")));
AggregationResults<BasicDBObject> groupResults = mongoTemplate.aggregate(
aggregation, "movieCollection", BasicDBObject.class);
return groupResults.getMappedResults();
}

Answer 2

$unwind of mongodb aggregation can be used for this.
db.Collection.aggregate([{
{$unwind : 'movies'},
{$match :{'movies.name' : 'fana'}}
}])
You can try the above query to get required output.

Answer 3

Above approaches provides you a solution using aggregation and basic query. But if you dont want to use BasicObject below code will perfectly work:
Query query = new Query()
query.fields().elemMatch("movies", Criteria.where("name").is("Fana"));
List<Movies> movies = mongoTemplate.find(query, Movies.class);
The drawback of this query is that it may return duplicate results present in different documents, since more than 1 document may match this criteria. So you can add _id in the criteria like below:
Criteria criteria = Criteria.where('_id').is(movieId)
Query query = new Query().addCriteria(criteria)
query.fields().elemMatch("movies", Criteria.where("name").is("Fana"));
query.fields().exclude('_id')
List<Movies> movies = mongoTemplate.find(query, Movies.class);
I am excluding "_id" of the document in the response.