在Mongo中检索子文档

Question

我是Mongo的新手并且正在努力检索特定文档的特定子文档

鉴于以下集合，我想检索function (a, b) { return a + b; }的内容。所以，我希望我的结果成为＆＃34; Java＆＃34;宾语。

"id" = 1 and "data.id" = 2

我正在执行{ "id" : "1", "title" : "Section1", "data" : [ { "id" : "1", "Product" : "Scala", "Description" : "", }, { "id" : "2", "Product" : "Java", "Description" : "", }, { "id" : "3", "Product" : "Ruby", "Description" : "", }, { "id" : "4", "Product" : "HTML5", "Description" : "", } ] } { "id" : "2", "title" : "Section2", "data" : [ { "id" : "4", "Product" : "Ansible", "Description" : "a free software platform for configuring and managing computers", }, { "id" : "1", "Product" : "Chef", "Description" : "an open source configuration management tool", }, { "id" : "2", "Product" : "Puppet", "Description" : "an open-source tool for managing the configuration of computer systems", }, { "id" : "3", "Product" : "Saltstack", "Description" : "open source configuration management and remote execution application", } ] } 之类的查询，但这会返回整个文档的ID为＆＃34; 1＆＃34;而不仅仅是我感兴趣的子文档。

我是否要求Mongo做一些不适合的事情，或者仅仅是我的语法有问题？

Answer 1

也许这会有所帮助：

db.myCollection.find({"id":"1","data.id":"2"},{"id":1,"data.$":1})

不需要传递$和

Answer 2

您可以使用聚合框架。

1. 检索匹配＆＃34;数据＆＃34;的文档项目。

   db.getCollection('yourColl').aggregate([ {$unwind:"$data"}, {$match:{"id":"1", "data.id":"2"}}, {$group:{_id:"$_id", "id":{$first:"$id"}, "title":{$first:"$title"}, "data":{$push:"$data"}}} ])

2. 如果您只想获取特定的子文档

   db.getCollection('yourColl').aggregate([ {$unwind:"$data"}, {$match:{"id":"1", "data.id":"2"}}, {$project:{_id:0, "id":"$data.id", "product":"$data.Product", "Description":"$data.Description"}} ])

Answer 3

您可以将$ match与$ unwind结合使用。这两个是聚合框架的一部分。

您的问题的答案是：

db.collection.aggregate([{$unwind:'$data'},{$match:{'data.Product':{$eq:'Java'}}}]);

参考：MongoDb unwind operator

Answer 4

回答了我自己的问题：

db.myCollection.find({$and : [{"id":"1"},{"data.id":"2"}]},{"data.$":1})