假设我有一个包含员工文档的MongoDB集合:{name:"JohnDoe",department:"Finance",salary:100}如何查询每个部门薪资最高的X名员工？编辑为了让自己更清楚一点，我是这样想的:db.collection.aggregate({$sort:{salary:-1}},{$group:{_id:"$department"employees:{$addToSet:"$name"}},{$project:{employees:{$slice:X}}})但这行不通有两个原因:1.$addToSet不保证输出集的任何顺序(至少根据documentation)
我希望使用mongoDB的聚合框架获得特定的查询。我想我需要$group和$addToSet运算符，但我对要使用的正确查询感到困惑。这是文章合集:/*0*/{"_id":4,"author":"KevinVanhove","book":{"order":500,"title":"HTML","url":"html"},"chapter":{"img":"navChapter-logo","order":500,"title":"W3C","url":"w3c"},"featured":0,"heading":[{"title":"title1","_id":ObjectId("5313
我有一个这样的文件集:{"Company":"4433","Descripcion":"trabajo","Referencia":"11817","HoraImportado":"15:54","ImportedOd":"2014-05-20T13:54:28.493Z","Items":[],"Notes":[{"_id":ObjectId("537b5ea4c61b1d1743f43420"),"NoteDateTime":"2014-05-20T13:54:44.418Z","Description":"nota","IsForTechnician":true,"Usernam
我需要对数组出现的次数求和。我需要将其输出到一个集合中，但是当我尝试使用$out关键字时，它失败并显示“不能将数组用于_id\”有没有办法将小组赛阶段的_id字段的值投影到一个新的key中，并创建一个新的_id？db.djnNews_filtered.aggregate([{$unwind:"$processed_text.headline_trigrams"},{$group:{_id:"$processed_text.headline_trigrams","num":{$sum:1}}},{$sort:{"num":-1}}]){"_id":["Reports","First","
以下代码始终返回相同的文档。req.query.skip一直递增10，仍然打印相同的文档集。还有1000多个文档。Collection.find({}).skip(req.query.skip).limit(10);varimageIdArray=[];cursor.each(function(err,item){if(item==null){//endconsole.log(imageIdArray);res.write(JSON.stringify(imageIdArray));res.end();}else{imageIdArray.push(item._id);}});skip
考虑数据集{"_id":{"$oid":"aaa"},"student_id":0,"type":"exam","score":54.6535436362647}{"_id":{"$oid":"bbb"},"student_id":0,"type":"quiz","score":31.95004496742112}{"_id":{"$oid":"ccc"},"student_id":0,"type":"homework","score":14.8504576811645}{"_id":{"$oid":"ddd"},"student_id":0,"type":"homework","sc
我问了Iscursor.skip()onindexedkeysalwaysfaster?现在我正在尝试利用multi-keyindexing(第94页)执行更快的查询。使用此脚本创建了2个数据库:vara=0;while(a++数据库索引如下:>db.fast.getIndexes()[{"v":1,"key":{"_id":1},"name":"_id_","ns":"test.fast"},{"v":1,"key":{"age":1},"name":"age_1","ns":"test.fast"},{"v":1,"key":{"age":1,"name":1,"email":1}
我有以下数据:{groupId:1,name:Jon},{groupId:1,name:Dan},{groupId:2,name:Jon},{groupId:2,name:Ris},{groupId:3,name:David}我收到一个groupID数组作为输入，我想计算这些组的DISTICT名称总数，我将聚合代码定义如下:groupIds[]={1,2}Aggregationagg=newAggregation(match(Criteria.where("groupId").in((groupIds)),group("name").count().as("total"));但我得到一
我是MongoDB的新手，所以如果我遗漏了文档中的某些内容，请原谅我。我的收藏是这样的:{"_id":ObjectId("57553e7015e4117a4343c18c"),"BuyingPrice":55.5,"Quantity":NumberLong(1),,"Brand":"Ranamina","Amount":79.99,"Profit":24.49,"ProductId":NumberLong(55319),}计算完每个品牌的总金额后，我想看看每个品牌有多少产品。我想看到结果{"Quantity":1982,"Amount":155,"Number_of_product":
我在MongoDB中有以下场景:每条记录都有自己的_id和parentId。如果parentId==""那么它就是一个真正的父记录。如果parentId有一个值，那么该记录实际上是一个指向父记录的子记录。下面显示了一个父项及其链接的子项。{"_id":ObjectId('586c9d275d2f62e1634978db'),parentId="",count=1,}{"_id":ObjectId('586c9d275d2f62e163497811'),parentId=ObjectId('586c9d275d2f62e1634978db'),count=3,}我想要一个查询，在那里我找
