MongoDB-分组依据-聚合-Java

Question

I have a doc in my mongodb that looks like this -

public class AppCheckInRequest {
    private String _id;
    private String uuid;
    private Date checkInDate;
    private Double lat;
    private Double lon;
    private Double altitude;
}

• The database will contain multiple documents with the same uuid but different checkInDates

Problem

I would like to run a mongo query using java that gives me one AppCheckInRequest doc(all fields) per uuid who's checkInDate is closest to the current time.

I believe I have to the aggregation framework, but I can't figure out how to get the results I need. Thanks.

Answer 1

In the mongo shell :-

This will give you the whole groupings:

db.items.aggregate({$group : {_id : "$uuid" , value : { $push : "$somevalue"}}} )

And using $first instead of $push will only put one from each (which is what you want i think?):

db.items.aggregate({$group : {_id : "$uuid" , value : { $first : "$somevalue"}}} )

Can you translate this to the Java api? or i'll try to add that too.

... ok, here's some Java:

Assuming the docs in my collection are {_id : "someid", name: "somename", value: "some value"} then this code shows them grouped by name:

 Mongo client = new Mongo("127.0.0.1");

    DBCollection col = client.getDB("ajs").getCollection("items");

    AggregationOutput agout = col.aggregate(
           new BasicDBObject("$group",
                   new BasicDBObject("_id", "$name").append("value", new BasicDBObject("$push", "$value"))));

    Iterator<DBObject> results = agout.results().iterator();


   while(results.hasNext()) {
     DBObject obj = results.next();

     System.out.println(obj.get("_id")+" "+obj.get("value"));

  }

and if you change $push to $first, you'll only get 1 per group. You can then add the rest of the fields once you get this query working.