我有一些文件{name:'apple',type:'fruit',color:'red'}{name:'banana',type:'fruit',color:'yellow'}{name:'orange',type:'fruit',color:'orange'}{name:'eggplant',type:'vege',color:'purple'}{name:'brocoli',type:'vege',color:'green'}{name:'rose',type:'flower',color:'red'}{name:'cauli',type:'vege',color:'white'}
我有以下格式的Assets文件。db.asset.find({}).limit(1).pretty(){"_id":ObjectId("54e650a10364a65f62c0df4a"),"_class":"com.model.core.Asset","displayName":"BingoRhymesForChildren+More3DAnimationNurseryRhymes&Kids'Songs","assetType":"VIDEO","active":true,"originalUrl":"https://www.youtube.com/watch?v=j-tdVvvXn
我正在尝试通过以下代码在mongoose中按聚合应用分组app.get('/api/get_conversation_all_carer/:carer_id',function(req,res){//Sender=userandreceiver=carervardata=[];varconversation={status_code:200,data:data};Conversation.find({'receiver_id':req.params.carer_id},{sender_id:1},{'group':'sender_id'},function(err,result){co
我必须使用mongoosegroupby和聚合函数来计算特定年份的数量，但我没有使用条件来对特定年份进行分组，我必须使用下面的代码，Opp.aggregate([{$project:{_id:{year:{$year:'$HireDate'}},hireDateValue:{year:{$year:'$HireDate'}},hireDateCount:{$cond:[{"$and":[{"$or":[{"$eq":["$HireDate","1994-01-1700:00:00.000"]}]}]},1,0]}}},{$group:{_id:'$hireDateValue',coun
我想计算每天的分配计数，通过在assigned_on上使用day分组我可以获得每天的分配计数，但那些是唯一计数每一天。在这一点上，我还希望包括昨天或几天前分配但尚未释放的那张床。比如我有如下记录{"assigned_on":ISODate("2015-12-01T00:00:00Z"),"released_on":ISODate("2015-12-01T14:01:23Z"),"bed_id":1},{"assigned_on":ISODate("2015-12-01T00:00:00Z"),"released_on":ISODate("2015-12-04T14:01:23Z"),"
我在mongoDB中有一组数据，我必须按$timestamp分组汇总。此字段包含日期，但格式为字符串(上面的示例数据)。我应该如何继续将$timestamp转换为日期以便将它们组合在一起？接下来，我必须对每个日期和标识的每个scores_today求和，每个scores_total也是如此。示例数据:[{_id:"1442",timestamp:"2016-03-15T22:24:02.000Z",iden:"15",scores_today:"0.000000",scores_total:"52337.000000"}]我的代码varproject={"$project":{"_id
我有一个听起来相当简单的任务，我想使用MongoDB的聚合管道来完成。我想将一个字段中的数组视为集合(即忽略顺序和重复项)，并按它们分组。例如，集合可能是:[{_id:1names:["a","b"]},{_id:2names:["c","a"]},{_id:3names:["b","a"]}]我想要返回的结果是这样的:[{names:["a","b"],count:2},{names:["a","c"],count:1}]谢谢! 最佳答案 您需要$sort结果使它们对于分组键保持一致。真的没有别的办法了:db.collection
我有这样的mongoDB集合{"_id":"EkKTRrpH4FY9AuRLj","stage":10,},{"_id":"EkKTRrpH4FY9AuRLj","stage":10,},{"_id":"EkKTRrpH4FY9AuRLj","stage":20,}我想把它分组在stage:10inpaidstage:20inunpaid结果应该是这样的{"paid":[{"_id":"EkKTRrpH4FY9AuRLj","stage":10,},{"_id":"EkKTRrpH4FY9AuRLj","stage":10,}],"unpaid":[{"_id":"EkKTRrpH4F
这是引用这个question.这是我的数据集:[{"rating":4,"ceatedAt":ISODate("2016-08-08T15:32:41.262+0000")},{"rating":3,"createdAt":ISODate("2016-08-08T15:32:41.262+0000")},{"rating":3,"ceatedAt":ISODate("2016-07-01T15:32:41.262+0000")},{"rating":5,"createdAt":ISODate("2016-07-01T15:32:41.262+0000")}]我希望能够在日期范围内以周或
在mongo中说我有一个看起来像这样的集合:+----+-----+-----+----------+|id|x|y|quantity|+----+-----+-----+----------+|1|abc|jkl|5|+----+-----+-----+----------+|2|jkl|xyz|10|+----+-----+-----+----------+|3|xyz|abc|20|+----+-----+-----+----------+我想做一个x等于y的$group并求和数量。所以输出看起来像:+-----+-------+|x|total|+-----+-------+|
