done_status=['BAD_PU','TO_WH','RCVDPORT','RCVD','BAD_DEL','MISSFLT','OFFLOAD']shipments=db.db_shipment.aggregate([{"$lookup":{"from":"db_shipment_status_history","localField":"_id","foreignField":"fk_shipment_id","as":"shipment_status_history_collection"}},{"$unwind":"$shipment_status_history_co
我正在尝试通过rake任务手动编辑“updated_at”字段这是它的样子:task:campaigns_updated_at_recovery=>:environmentdoDir.foreach('db/raw-data/campaigns/')do|json|nextifjson=='.'orjson=='..'file=File.read('db/raw-data/campaigns/'+json)data_hash=JSON.parse(file)#pdata_hash["_id"]thisCampaign=Campaign.find(data_hash["_id"])thi
我有两个名为博客类别和SubBlogCategory。BlogCategory的文档由子类别id的objectId数组组成。类目文档如下{"_id":ObjectId("5af2c6e8bfab7269e9a42ded"),"title":"Javascript","subcategory":[ObjectId("5af29fcc9a52623b7088ef4e"),ObjectId("5aebf78681273424e5f55ecc")]}子分类文档如下{"_id":ObjectId("5af29fcc9a52623b7088ef4e"),"title":"Reactjs"}{"_i
我有一个channel收藏{channel:"Xamper",//uniquesubscribers:[ObjectId("5a934e000102030405000000"),ObjectId("5a934e000102030405000001"),ObjectId("5a934e000102030405000002")]}还有一个用户集合{_id:ObjectId("5a934e000102030405000000"),name:"Bradman"},{_id:ObjectId("5a934e000102030405000001"),name:"Hartin"},{_id:Obje
我正在考虑使用最佳实践来解决Mongoose的以下问题。我有三个模式:constSchemaA=newSchema({name:String});constSchemaB=newSchema({schemaA:{type:Schema.Types.ObjectId,ref:'SchemaA',required:true}});constSchemaC=newSchema({schemaB:{type:Schema.Types.ObjectId,ref:'SchemaB'},user:{type:Schema.Types.ObjectId,ref:'User'}});我需要通过附加了sc
我正在尝试制作一个聚合管道-$lookup以从另一个接收仅收集不等于特定_id的项目例如:诊所集合:{_id:1,name:'somename1'}{_id:2,name:'somename2'}{_id:3,name:'somename3'}业务集合:{_id:1,name:"somebusinessname",clinics:[1,2,3]}我的聚合管道查询:db.business.aggregate([{$match:{_id:mongoose.Types.ObjectId(businessId)}},{$lookup:{from:"ClinicsCollection",loca
我有两个集合，并试图在两个不同的条件下得到总金额的总和。这是我的代码:db.drivers.aggregate([{$lookup:{from:"bookings",let:{driver_id:"$_id"},pipeline:[{$match:{$expr:{$eq:["$driverId","$$driver_id"]}}},{$group:{_id:"$driverId",totalAmount:{$sum:"$driverAmount"}}}],as:"bookingdata",pipeline:[{$match:{$expr:{$and:[{$eq:["$driverId"
我有3个收集司机、用户和调度员。调度员可以是司机或用户我的调度员集合是这样的，{"_id":ObjectId("5c9bb56a56f1a43e8aedf3c1"),"dispatcherId":ObjectId("5c3382fe3406281ba6863f7e"),"type":"driver","recordedTime":ISODate("2019-03-27T17:39:54.030Z"),"isEnable":true,"dispatchPackageType":"commission","__v":0}dispatcherId来自驱动程序或用户集合。类型是“驱动程序”或“
我循环浏览我收藏的文档，做一些事情，然后更新数据库。但是，由于我实际上拥有正在更新的文档的所有数据，如果我这样做，save()会比update()更快吗？foreach($cursoras$doc){$doc['new_field']='value';$coll->save($doc);/*or(currently)*/$coll->update(array('known_field'=>$doc['known_field']),array('$set'=>array('new_field'=>'value')));}哪种方式更快？ 最佳答案
我如何将一个项目推送到数组中？我发现我只能插入基本值(String、Int32、Int64、Boolean)，但我不能将自定义类的实例插入到数组中。//inthisway,itwork:varmyPlayer=newi_Player();this.mongo_collection.FindAndModify(Query.EQ("_id",ID),SortBy.Ascending("_id"),Update.PushWrapped("_player",myPlayer),true);//inthisway,don'tworkbecausei_PlayerisnotanBsonValueb
