我正在尝试从MongoDB中获取数据，以获得前5个最受欢迎的产品 . 但我收到了以下错误 . （MongoDB版本3.2.21）com.mongodb.MongoCommandException：命令失败，错误9：''cursor'选项是必需的，除了在服务器localhost：27017上使用explain参数的聚合 . 完整的响应是{“ok”：0.0，“errmsg”：“'cursor'选项是必需的，除了带有explain参数的聚合”，“code”：9，“codeName”：“FailedToParse”}

DBObject groupFields = groupFields = new BasicDBObject("_id", 0);
groupFields.put("_id", "$productId");               
groupFields.put("avgRating", new BasicDBObject("$avg", "$productRvR")); 
groupFields.put("productModelName", new BasicDBObject("$push", "$productMN"));
DBObject group = new BasicDBObject("$group", groupFields);

DBObject projectFields = new BasicDBObject("_id", 0);
projectFields.put("productId", "$_id");
projectFields.put("avgRating", "$avgRating");
projectFields.put("productModelName", "$productModelName");
DBObject project = new BasicDBObject("$project", projectFields);

DBObject sort = new BasicDBObject();
sort.put("avgRating",-1);
DBObject orderby=new BasicDBObject("$sort",sort);
DBObject limit=new BasicDBObject("$limit",5);

AggregationOutput aggregate userReviews.aggregate(group,project,orderby,limit);

LinkedHashMap<String, String> mostLikedProductsMap = new LinkedHashMap<String, String>();

for (DBObject queryResult : aggregate.results()) {
    BasicDBObject obj = (BasicDBObject) queryResult;
    BasicDBList productModelName = (BasicDBList) obj.get("productModelName");

    mostLikedProductsMap.put((String)productModelName.get(0),obj.getString("avgRating"));

                }

return mostLikedProductsMap;
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