如何从所有文档中仅返回数组的嵌套文档 [英] How to return just the nested documents of an array from all documents

问题描述

我有一个关于查询嵌套文档的问题.我试图搜索，但没有任何答案回答我，或者我可能忽略了它.我有这样的结构:

I have a question about querying nested documents. I tried to search but nothing answered my question or I am maybe overlooking it. I have structure like this:

    {
    "_id" : ObjectId("592aa441e0f8de09b0912fe9"),
    "name" : "Patrick Rothfuss",
    "books" : [ 
    {
        "title" : "Name of the wind",
        "pages" : 400,
        "_id" : ObjectId("592aa441e0f8de09b0912fea")
    }, 
    {
        "title" : "Wise Man's Fear",
        "pages" : 500,
        "_id" : ObjectId("592aa441e0f8de09b0912feb")
    },
    },
    {
    "_id" : ObjectId("592aa441e0f8de09b0912fe9"),
    "name" : "Rober Jordan",
    "books" : [ 
    {
        "title" : "The Eye of the World",
        "pages" : 400,
        "_id" : ObjectId("592aa441e0f8de09b0912fea")
    }, 
    {
        "title" : "The Great Hunt",
        "pages" : 500,
        "_id" : ObjectId("592aa441e0f8de09b0912feb")
    }
    },

我想查询所有作者中所有图书的列表-类似于:

And I would like to query for the list of all books in entire colletion of Authors - something like:

"books" : [ 
    {
        "title" : "The Eye of the World",
        "pages" : 400,
        "_id" : ObjectId("592aa441e0f8de09b0912fea")
    }, 
    {
        "title" : "The Great Hunt",
        "pages" : 500,
        "_id" : ObjectId("592aa441e0f8de09b0912feb")
    },
    {
        "title" : "Name of the wind",
        "pages" : 400,
        "_id" : ObjectId("592aa441e0f8de09b0912fea")
    },
    {
        "title" : "Wise Man's Fear",
        "pages" : 500,
        "_id" : ObjectId("592aa441e0f8de09b0912fea")
    }]

推荐答案

您可以使用 .aggregate() ，主要是 $unwind 管道运算符:

You can do this using .aggregate() and predominantly the $unwind pipeline operator:

在现代MongoDB 3.4及更高版本中，您可以与

In modern MongoDB 3.4 and above you can use in tandem with $replaceRoot

Model.aggregate([
  { "$unwind": "$books" },
  { "$replaceRoot": { "newRoot": "$books" } }
],function(err,results) {

})

在早期版本中，您使用 $project :

In earlier versions you specify all fields with $project:

Model.aggregate([
  { "$unwind": "$books" },
  { "$project": {
    "_id": "$books._id",
    "pages": "$books.pages",
    "title": "$books.title"
  }}
],function(err,results) {

})

所以 $unwind 是您用来解构的内容或反规范化"数组条目以进行处理.有效地，这为数组的每个成员创建了整个文档的副本.

So $unwind is what you use to deconstruct or "denormalise" the array entries for processing. Effectively this creates a copy of the whole document for each member of the array.

剩下的任务是关于仅"返回数组中存在的那些字段.

The rest of the task is about returning "only" those fields present in the array.

这不是一件很明智的事情.如果您打算只返回嵌入在文档数组中的内容，那么最好将这些内容放入单独的集合中.

It's not a very wise thing to do though. If your intent is to only return content embedded within an array of a document, then you would be better off putting that content into a separate collection instead.

在性能上要好得多，使用聚合框架将集合中的所有文档分开，仅列出数组中的那些文档即可.

It's far better for performance, pulling apart a all documents from a collection with the aggregation framework, just to list those documents from the array only.

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