How to read only a particular JSON node from MongoDb in SpringBoot

Question

I'm trying to read a single JSON node from MongoDb via my Springboot project.

here is the sample JSON saved in Mongo:

    "a": {
            "b": {
                "c": {
                    "d": "value1",
                    "e": "value2"
                }
            }
        }

I tried:


   public String get() {
       BasicDBObject query = BasicDBObject.parse("{\"a.b.c.d\": \"value1\"}");
       FindIterable<Document> dumps = mongoOperations.getCollection("data").find(query);
       return dumps.first().toJson();
   }

response I got:


    {
        "_id": {
            "$oid": "5e8aa42602ef9f05bf35ff59"
        },
        "a": {
            "b": {
                "c": {
                    "d": "value1",
                    "e": "value2"
                }
            }
        }
    }

but I only need the node a.b.c.d, I know I can do it in java after reading the whole JSON. But my intention is to read only the needed part.

Expected Output:


    {
       "a": {
            "b": {
                "c": {
                    "d": "value1"
                }
            }
        }
    }

Also please note that I'll be saving JSONs with random schema so no POJOs associated with it. It would be really helpful if somebody can guide me on this.

prasad_ · Accepted Answer · 2020-04-06 06:47:54Z

1

You can get the specified field only, using a projection with the find query. Tried with MongoDB Spring Data 2.2.6:

MongoOperations mongoOps = new MongoTemplate(MongoClients.create(), "testDB");

Query qry = new Query();
qry.fields().include("a.b.c.d").exclude("_id");
List<Document> list = mongoOps.find(qry, Document.class, "collection");
list.forEach(doc -> System.out.println(doc.toJson()));

The output:

{"a": {"b": {"c": {"d": "value1"}}}}

answered Apr 6, 2020 at 6:47

prasad_

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