Spring Mongo聚合按carId分组取min值并保留所有字段的方案

需要获取每个carId对应车辆的最快速度（即raceResulttime最小）的完整记录。当前使用Spring Mongo聚合框架按carId分组并取raceResulttime最小值后，返回结果仅保留carId和raceResulttime，其他字段均为null。尝试过的方案要么将字段存入数组，要么无法保证carId唯一。

集合数据 #

Race{raceId='64c2350402708c36b821e5bc', raceNum=430, trackNum='2', carUser='null', carId='ZZ9992', raceStart=null, raceEnd=null, raceResulttime=2.262371, raceDatetime=Thu Jul 27 17:12:36 HKT 2023, venue='HK', rank='2', isDnf=false, raceSpeed=0.0}, 
Race{raceId='64c2355f02708c36b821e5bf', raceNum=430, trackNum='2', carUser='null', carId='ZZ9992', raceStart=null, raceEnd=null, raceResulttime=2.262371, raceDatetime=Thu Jul 27 17:14:07 HKT 2023, venue='HK', rank='2', isDnf=false, raceSpeed=0.0}, 
Race{raceId='64c2357002708c36b821e5c2', raceNum=431, trackNum='2', carUser='null', carId='ZZ9992', raceStart=null, raceEnd=null, raceResulttime=2.262371, raceDatetime=Thu Jul 27 17:14:24 HKT 2023, venue='HK', rank='2', isDnf=false, raceSpeed=0.0}, 
Race{raceId='64c2357e02708c36b821e5c5', raceNum=432, trackNum='2', carUser='null', carId='ZZ9992', raceStart=null, raceEnd=null, raceResulttime=2.262371, raceDatetime=Thu Jul 27 17:14:38 HKT 2023, venue='HK', rank='2', isDnf=false, raceSpeed=0.0}, 
Race{raceId='64c2358d02708c36b821e5c8', raceNum=433, trackNum='2', carUser='null', carId='ZZ9992', raceStart=null, raceEnd=null, raceResulttime=2.262371, raceDatetime=Thu Jul 27 17:14:53 HKT 2023, venue='HK', rank='2', isDnf=false, raceSpeed=0.0}, 
Race{raceId='64c23282bc05ab7f8acd7c2a', raceNum=421, trackNum='1', carUser='null', carId='ZZ9991', raceStart=null, raceEnd=null, raceResulttime=4.298986, raceDatetime=Thu Jul 27 17:01:54 HKT 2023, venue='HK', rank='3', isDnf=false, raceSpeed=0.8806175}, 
Race{raceId='64c23294bc05ab7f8acd7c2c', raceNum=422, trackNum='1', carUser='null', carId='ZZ9991', raceStart=null, raceEnd=null, raceResulttime=4.298986, raceDatetime=Thu Jul 27 17:02:12 HKT 2023, venue='HK', rank='3', isDnf=false, raceSpeed=0.8806175}

现有聚合代码 #

@Override
public void groupByField() {
    // Match stage
    MatchOperation matchStage = Aggregation.match(new Criteria("raceResulttime").gt(0)
            .and("isDnf").is(false)
            .and("raceNum").gt(0)
    );
    GroupOperation groupStage = group("carId")
            .min("raceResulttime").as("raceResulttime");
    SortOperation sortStage = Aggregation.sort(Sort.Direction.ASC, "raceResulttime");
    ProjectionOperation projectionOperation = Aggregation.project("raceResulttime","rank","raceSpeed","raceNum").and("carId").previousOperation();
    // Aggregation
    Aggregation aggregation = newAggregation(matchStage, groupStage, sortStage,projectionOperation);
    AggregationResults<Race> resultss = mongoTemplate.aggregate(aggregation, "race", Race.class);
    for ( Race result : resultss)
        System.out.println("4 - resultss - "+ result);
    System.out.println("5" + resultss.getMappedResults());
}

当前返回结果 #

- resultss - Race{raceId='null', raceNum=null, trackNum='null', carUser='null', carId='ZZ9992', raceStart=null, raceEnd=null, raceResulttime=1.243837, raceDatetime=null, venue='null', rank='null', isDnf=false, raceSpeed=null}
4 - resultss - Race{raceId='null', raceNum=null, trackNum='null', carUser='null', carId='ZZ9991', raceStart=null, raceEnd=null, raceResulttime=1.391792, raceDatetime=null, venue='null', rank='null', isDnf=false, raceSpeed=null}
4 - resultss - Race{raceId='null', raceNum=null, trackNum='null', carUser='null', carId='ZZ9990', raceStart=null, raceEnd=null, raceResulttime=1.483507, raceDatetime=null, venue='null', rank='null', isDnf=false, raceSpeed=null}
4 - resultss - Race{raceId='null', raceNum=null, trackNum='null', carUser='null', carId='ZZ9995', raceStart=null, raceEnd=null, raceResulttime=1.493411, raceDatetime=null, venue='null', rank='null', isDnf=false, raceSpeed=null}
4 - resultss - Race{raceId='null', raceNum=null, trackNum='null', carUser='null', carId='ZZ9996', raceStart=null, raceEnd=null, raceResulttime=1.661812, raceDatetime=null, venue='null', rank='null', isDnf=false, raceSpeed=null}

核心思路：先按carId和raceResulttime升序排序，确保每组内最快的记录排在首位；再按carId分组，取每组第一条完整记录，即可保留该车辆最快速度对应的所有字段。

修改后的聚合代码 #

@Override
public void groupByField() {
    // 过滤有效记录
    MatchOperation matchStage = Aggregation.match(new Criteria("raceResulttime").gt(0)
            .and("isDnf").is(false)
            .and("raceNum").gt(0)
    );
    
    // 先按carId分组排序，再按raceResulttime升序（最快记录排前面）
    SortOperation sortStage = Aggregation.sort(Sort.Direction.ASC, "carId", "raceResulttime");
    
    // 按carId分组，取每组第一条完整记录
    GroupOperation groupStage = Aggregation.group("carId")
            .first(Aggregation.ROOT).as("fastestRace");
    
    // 提取完整的Race对象，排除自动生成的_id
    ProjectionOperation projectionStage = Aggregation.project("fastestRace")
            .andExclude("_id");
    
    // 构建聚合管道
    Aggregation aggregation = Aggregation.newAggregation(matchStage, sortStage, groupStage, projectionStage);
    
    // 执行聚合，先获取Document结果
    AggregationResults<Document> results = mongoTemplate.aggregate(aggregation, "race", Document.class);
    
    // 将Document转换为Race对象
    List<Race> fastestRaces = results.getMappedResults().stream()
            .map(doc -> mongoTemplate.getConverter().read(Race.class, doc.get("fastestRace", Document.class)))
            .collect(Collectors.toList());
    
    // 输出结果
    for (Race race : fastestRaces) {
        System.out.println("最快记录：" + race);
    }
}

效果说明 #

修改后，每个carId仅返回一条记录，包含该车辆最快速度对应的所有字段（raceId、raceNum、trackNum等），且保证carId唯一，完全符合需求。

内容的提问来源于stack exchange，提问作者Man Man Yu