运行以下查询时，结果显示在mongoshell中。示例记录{"_id":ObjectId("587e21df6e79d255011a9c6a"),"vendor_id":"101","subscription_id":14,"created_at":ISODate("2017-01-17T13:53:35.272Z")}{"_id":ObjectId("587e21df6e79d255011a9c6c"),"vendor_id":"102","subscription_id":14,"created_at":ISODate("2017-01-17T13:56:35.272Z")}查询d
我有一个mongodb数据库，其中包含看起来像这样的公司集合(这只是一个示例，实际集合要大得多):[{"_id":100,"name":"TestName1","level":"1"},{"_id":101,"name":"TestName2","level":"1"},{"_id":102,"name":"TestName3","level":"2"}]其中“level”的范围只能是0到5我正在尝试使用$group和$project进行聚合查询，计算每个级别中有多少家公司，但根据我需要遵循的API规范，它需要像这样格式化，在一个单一的对象:{"metrics":{"companies
我想在$bucket函数中单独处理$group的结果。我的数据:{"_id":1,"title":"ThePillarsofSociety","artist":"Grosz","year":1926,"type":1}{"_id":2,"title":"MelancholyIII","artist":"Munch","year":1902,"type":1}{"_id":3,"title":"Dancer","artist":"Miro","year":1925,"type":1}{"_id":4,"title":"TheGreatWaveoffKanagawa","artist":"
我有以下文档结构{"timestamp":13512493603565120,"value":1,"y":42,"M":513,"w":2234,"d":15639,"S":46918,"h":375347,"m":22520822,"s":1351249360,"_id":ObjectId("508aa61100b5457c04000001"),"__v":0}我有一个如下的mongodb聚合来汇总按字段y分组的值:aggregate({$group:{_id:"$y",value:{$sum:4}}})这会给我[{"_id":42,"value":16}]我现在想要的是格式化此输出
假设我有一个包含员工文档的MongoDB集合:{name:"JohnDoe",department:"Finance",salary:100}如何查询每个部门薪资最高的X名员工？编辑为了让自己更清楚一点，我是这样想的:db.collection.aggregate({$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","
我正在玩弄SpringDataRest。我无法完成的一件事是将嵌套对象存储在专用存储库中。这是我的两个模型类Person和Address:@EntitypublicclassAddress{@NotEmptypublicStringaddress,email;@IdpublicStringid;}@EntitypublicclassPerson{@IdpublicStringid;publicStringfirstName,lastName;@OneToOnepublicAddressaddress;}这是我在SpringBoot应用程序中使用的两个Mongo存储库。@Reposito
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
