下面是示例文档结构。
db.volume_statistics.insert({ "client": "SER",
"message": "Sample_v02RQ",
"GDS": "SABRE",
"daily": 240000,
"hourly": {"0": 1000, "1": 100, "2": 454, "3":3434, "4":3434, "5":343, ... , ”23”:454 },
"date":ISODate("2013-06-23T04:00:00Z") });我正在尝试编写一个查询,按照下面的输出格式按8小时{所有8小时的总和}和月份对每小时列的信息进行分组。
{ "Month" : 5 , "01-08" : 26970, "09-17" : 45970}
{ "Month" : 6 , "01-08" : 269712, "09-17" : 56970}输出定义了6月{6}月,01-08小时内收到269712个请求,09-17小时内收到56970个请求。
我已经编写了以下本机命令和Java代码来执行此操作。
本机命令::
db.volume_statistics.aggregate( { $match: { date: { $gt: ISODate("2011-01-01T00:00:00Z") } } },
{ $group : { _id: { month: { $month: "$date" } }, count: { $sum: "$hourly.0" }, count1: { $sum: "$hourly.1" } } },
{ $project: { _id: 1, count : 1 , count1 : 1 , "01-08" :{ $add:["$count", "$count1"]} } } );Java实现:
DBObject match = new BasicDBObject("$match", new BasicDBObject("date",new BasicDBObject("$gt",fromDate)));
DBObject fields = new BasicDBObject("_id", 1);
fields.put("hourly_0", 1);
fields.put("hourly_1", 1);
fields.put("01-08", new BasicDBObject("$add","$hourly_0"));
DBObject project = new BasicDBObject("$project", fields);//现在$group操作*
DBObject groupFields = new BasicDBObject("_id", new BasicDBObject("$month","$date"));
groupFields.put("hourly_0", new BasicDBObject("$sum", "$hourly.0"));
groupFields.put("hourly_1", new BasicDBObject("$sum", "$hourly.1"));
DBObject group = new BasicDBObject("$group", groupFields);*
//运行聚合
AggregationOutput output = table.aggregate(match,group,project);我几乎已经完成了逻辑,但$add命令的唯一问题。如何使用mongo-java实现定义“$add :"$count”,“$count1”“。我不能在java实现的$add命令中提到多个值。有人能建议我如何完成这项工作吗?
发布于 2013-06-28 00:27:01
在Java语言中使用BasicDBList来表示数组。您的代码将如下所示:
DBObject fields = new BasicDBObject("_id", 1);
fields.put("hourly_0", 1);
fields.put("hourly_1", 1);
BasicDBList dbList = new BasicDBList();
dbList.add("$hourly_0");
dbList.add("$hourly_1");
fields.put("01-08", new BasicDBObject("$add",dbList));
DBObject project = new BasicDBObject("$project", fields);发布于 2013-09-19 02:55:38
如何通过mongo-java实现定义“$add :"$count”,“$count1”“
String [] srt_array = {"$count", "$count1"};
BasicDBObject add_obj = BasicDBObject ("$add", srt_array);https://stackoverflow.com/questions/17335037
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