Question

示例日期（集合名称＆＃39;人员＆＃39;）

{"person" : "Frank", 
 "coords" : [
    {
        "point":new GLatLng(40.266044,-74.718479), 
    },
    {
        "point":new GLatLng(40.211600,-74.695702),
    }
] }

{"person : ""Bob", 
  "coords" : [
        {
            "point":new GLatLng(40.266044,-74.718479), 
        },
        {
            "point":new GLatLng(40.211600,-74.695702),
        },
        {
            "point":new GLatLng(40.294535,-74.682012),
        }
] }

目标：获得＆＃39; coord＆＃39;的总数。每个人文件的条目（总体累计）

预期结果5

我尝试了什么

public int getTotalCoordinateCount(){

  BasicDbObject project= new BasicDBObject()
  .append("count",new BasicDBObject("$size","coords"))

   AggregationOutput output = db.getPeopleCollection().aggregate(Arrays. <DBObject>asList(
     new BasicDBObject("$project", project);

    BasicDbObject result = (BasicDBObject)output.results().iterator.next();

   return result.getInt("count");

}

问题我得到：2

对我在mongo聚合框架中做错的事情一无所知。我怀疑我需要包含$sum，但不确定

Answer 1

您需要在聚合管道中的coords字段上执行展开操作。之后，您按Null并使用count字段进行分组，以计算结果集，您将获得所有文档中所有数组的元素总数。

Answer 2

public int getTotalCoordinateCount(){

  BasicDbObject group = new BasicDBObject()
  .append("_id", 1)
  .append("count", new BasicDBObject("$sum", 1));

  BasicDbObject project= new BasicDBObject()
  .append("_id", 0)
  .append("count", 1);

   AggregationOutput output = db.getPeopleCollection().aggregate(Arrays. <DBObject>asList(
     new BasicDBObject("$unwind", "coords"),
     new BasicDBObject("$group", group),
     new BasicDBObject("$project", project)
   );

    BasicDbObject result = (BasicDBObject)output.results().iterator.next();

   return result.getInt("count");

}

Mongo：计算集合中每个条目的数组元素数

2 个答案: