Tìm tài liệu nhúng riêng biệt và tạo sự khác biệt hơn nữa bằng cách sử dụng Trường

Bạn có thể sử dụng tập hợp MongoBD để có được kết quả mong muốn (Hãy xem xét ):

db.health.aggregate([
  {
    $sort: {
      "healths.effDate": 1
    }
  },
  {
    $group: {
      _id: "$healths.healthCd",
      healths: {
        $first: "$healths"
      }
    }
  },
  {
    $replaceRoot: {
      newRoot: "$healths"
    }
  }
])

MongoPlayground

Triển khai khởi động mùa xuân

package com.example.demo;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.boot.CommandLineRunner;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.data.domain.Sort.Direction;
import org.springframework.data.mongodb.core.MongoTemplate;
import org.springframework.data.mongodb.core.aggregation.Aggregation;
import org.springframework.data.mongodb.core.aggregation.AggregationResults;

import com.google.gson.Gson;
import com.google.gson.GsonBuilder;

@SpringBootApplication
public class DemoApplication implements CommandLineRunner {

    @Autowired
    private MongoTemplate mongoTemplate;

    public static void main(String[] args) {
        SpringApplication.run(DemoApplication.class, args);
    }

    @Override
    public void run(String... args) throws Exception {

//      //If your operator is not available inside Aggregation or query is too complex, 
//      //use below code to write MongoDB shell code directly as JSON
//      new AggregationOperation() {
//
//          @Override
//          public Document toDocument(AggregationOperationContext context) {
//              return new Document("$group", 
//                  new Document("_id", "$healths.healthCd")
//                      .append("healths", new Document("$first", "$healths")));
//          }
//          
//      },

        Aggregation agg = Aggregation.newAggregation(
            Aggregation.sort(Direction.ASC, "healths.effDate"),
            Aggregation.group("healths.healthCd").first("healths").as("healths"),           
            Aggregation.replaceRoot("healths")
        );

        AggregationResults<Healths> healths = mongoTemplate.aggregate(agg, 
                mongoTemplate.getCollectionName(Health.class), Healths.class);
        for (Healths health : healths.getMappedResults()) {
            Gson gson = new GsonBuilder().setPrettyPrinting().create();
            System.out.println(gson.toJson(health));
        }
    }
}