想象一下mongodb上的以下集合:{id:1,array1:[ObjectId("1001"),ObjectId("1002")],array2:[ObjectId("1003"),ObjectId("1004")]}{id:2,array1:[ObjectId("1001"),ObjectId("1004")],array2:[ObjectId("1002")]}{id:3,array1:[ObjectId("1003"),ObjectId("1004")],array2:[ObjectId("1005")]}{id:4}现在我想创建一个这样的投影:db.coll.aggregat
我想做一个聊天系统，还需要获取用户的最后一条消息。我也提供了查询，但它只返回用户的userId。所以请帮助我，谢谢数据库:/*1*/{"_id":ObjectId("56937df0418a6afab248616d"),"to":ObjectId("56728051d4b426be03de18f2"),"from":ObjectId("568e402eaecfa53282f60d17"),"msg":"Hello!","cd":ISODate("2016-01-11T10:03:28.139Z"),"type":"other","ir":0}/*2*/{"_id":ObjectId("
这个问题在这里已经有了答案:MongoDB:aggregationframework:$matchbetweenfields(2个答案)关闭6年前。所以我有这个查询，db.collection.find({$where:"this.field1!==this.field2"})但现在我需要创建一个类似的查询并将结果聚合到一个经过尝试且真实的复杂查询中，只能通过使用聚合管道或“大炮飞”并使用mapReduce选项来完成。因为我想避免使用mapReduce，有没有办法实现类似于{$where:"this.field1!==this.field2"}方法？一些观察，解决上述情况的一般方法的答
这是我的数据库。{"_id":ObjectId("579cab6c6aba30f42a57a979"),"iecode":"P1111","country":"India","totalTreatmentArms":3,"treatmentArms":[{"mechanismOrPkg":"Mechanism","mechanism":"mechanism1"},{"mechanismOrPkg":"Mechanism","mechanism":"mechanism2"},{"mechanismOrPkg":"Package","mechanism":"mechanism1"}]}{"
示例数据集:{"source":"http://adress.com/","date":ISODate("2016-08-31T08:41:00.000Z"),"author":"SomeGuy","thread":NumberInt(115265),"commentID":NumberInt(2693454),"title":["A","title","for","a","comment"],"comment":["This","is","a","comment","with","a","duplicate"]}我使用的数据集基本上是来自用户的评论，具有唯一的commentID。评论
当我对我的收藏执行此查询时...models.Project.find(function(err,result){//result=docbelow}).populate('media')...我得到这个结果:{_id:57f36baa6cf34d079c8474a0,code:'ZMIA',__v:0,media:[{_id:57f36bb26cf34d079c847766,project_code:'ZMIA',amount:228,__v:0},{_id:57f36bb26cf34d079c84775c,project_code:'ZMIA',amount:250,__v:0}]
以下查询:[{$match:{$and:[{campaignID:"12345"},{date:"2016-10-18"},{hour:{$gte:0}},{hour:{$lte:3}}]}}}]返回以下(正确的)数据:[{"_id":"580638b1e4b01c1a027e82c1","campaignID":"12345","date":"2016-10-18","hour":0,"foo":10,"bar":10},{"_id":"580638cbe4b01c1a027e82c2","campaignID":"12345","date":"2016-10-18","hour":
假设我有这个文档:{"_id":1,"first_name":"Matt","roles":['editor','publisher','siteAdmin']}{"_id":1,"first_name":"Harry","roles":['publisher','siteAdmin']}{"_id":1,"first_name":"Rob","roles":['editor']}现在我想使用mongoose聚合($cond)来输出这个文档。我怎么得到这个？非常感谢您的帮助。{"first_name":"Matt","isSiteAdmin":true}{"first_name":"H
我正在尝试计算给定用户有多少条未读通知。也就是说，通过aggregate，我已经在控制台中完成了输出预期结果的下一个代码:db.getCollection('users').aggregate([{$match:{_id:ObjectId("5847f61a825d024ac9e3d08c")}},{$unwind:'$notifications'},{$match:{'notifications.read':{$eq:0}}},{$group:{_id:'$_id',notifications:{$push:'$notifications._id'}}}])哪些输出:{"_id":O
我有一个收藏:{_id:xxx,children:[{childrenOfChildren:[{price:xxx},{price:xxx},{price:xxx}]},{childrenOfChildren:[{price:xxx},{price:xxx},{price:xxx}]},{childrenOfChildren:[{price:xxx},{price:xxx},{price:xxx}]}]},{_id:xxx,children:[{childrenOfChildren:[{price:xxx},{price:xxx},{price:xxx}]},{childrenOfCh
