使用Redis从存储集中获取所有唯一分数？

问题描述

要添加地理数据，请使用以下代码

jedis.geoadd("storegeodata",51.5074,0.1278,"London");
//while member string is mostly json 
jedis.geoadd("storegeodata","{place: London}");
jedis.geoadd("storegeodata","{place: London,lat: 51.5074,lon: 0.1278}");

由于成员字符串不同而重复了geohash时，如何获取基于geohash的唯一值以避免冗余

public Map<String,Object> checkRedisGeo(double lat,double lon,Jedis j) {
    Map<String,Object> result = new HashMap<>();
    try
    {
        GeoRadiusParam param = GeoRadiusParam.geoRadiusparam();
        param.sortAscending();
        param.withdist();
        param.count(1);
        System.out.println("lat :"+lat+",lon :"+lon);
        List<GeoRadiusResponse> response =  j.georadius("commander",lon,lat,150,GeoUnit.M,param);    
        //System.out.println(response.size()+" size");
      if(response.size() > 0)
      {
          for (GeoRadiusResponse geoRadiusResponse : response) {
                System.out.println("lat :"+lat+",lon :"+lon+",stringmember :"+geoRadiusResponse.getMemberByString());
                //System.out.println(geoRadiusResponse.getdistance());
                Object[] data= {geoRadiusResponse.getMemberByString()};
                System.out.println(data);
                result.put("result",data);
        
               }
      }else {
         // sendEvents(streamEvent,null,streamEventChunk);
          System.out.println("E");
      }
      
    } catch (Exception e) {
        LOGGER.error("checkRedisGeo err : "+e);
    }
      
  return result;
    
}

哪个会检索结果，但是如何根据geohash / score值进行过滤，如何获取所有不同的分数？

以下是示例冗余数据

解决方法

由于得分是唯一的，因此能够通过比较得分来获得重复和唯一得分 并按照相同的代码存储了多个相同得分的成员

    Double previous_score = 0.0;
    int DuplincatesCount = 0;
    int UniqueCount = 0;
    Set<String> values = jedis.zrange("rediskey",-1);// get all the members
    for (String member : values) {
        Double current_score = jedis.zscore("rediskey",member);//each member looped
        if (Double.compare(current_score,previous_score) == 0) { //comparing current score with previous
                                                                    // score
            DuplincatesCount++;
        } else {
            UniqueCount++;
        }
        previous_score = current_score;// score mapping
    }
    System.out.println("Duplincates entry " + DuplincatesCount);
    System.out.println("Unique entry " + UniqueCount);