我如何在MongoDB中执行SQL Join等效?

例如,假设你有两个集合(用户和评论),我想拉pid=444的所有评论以及每个评论的用户信息。

comments
  { uid:12345, pid:444, comment="blah" }
  { uid:12345, pid:888, comment="asdf" }
  { uid:99999, pid:444, comment="qwer" }

users
  { uid:12345, name:"john" }
  { uid:99999, name:"mia"  }

是否有一种方法可以一次性提取具有特定字段的所有评论(例如. ...find({pid:444}))和与每个评论相关的用户信息?

目前,我首先获取符合条件的注释,然后找出结果集中的所有uid,获取用户对象,并将它们与注释的结果合并。看来我做错了。


当前回答

您可以使用聚合管道来实现它,但是自己编写它很麻烦。

您可以使用mongo-join-query从您的查询自动创建聚合管道。

这是你的查询的样子:

const mongoose = require("mongoose");
const joinQuery = require("mongo-join-query");

joinQuery(
    mongoose.models.Comment,
    {
        find: { pid:444 },
        populate: ["uid"]
    },
    (err, res) => (err ? console.log("Error:", err) : console.log("Success:", res.results))
);

您的结果将在uid字段中有user对象,您可以链接任意多的层次。您可以填充对用户的引用,从而引用一个Team,再引用其他东西,等等。

免责声明:我编写了mongo-join-query来解决这个问题。

其他回答

在3.2.6之前,Mongodb不像mysql那样支持join查询。下面是适合你的解决方案。

 db.getCollection('comments').aggregate([
        {$match : {pid : 444}},
        {$lookup: {from: "users",localField: "uid",foreignField: "uid",as: "userData"}},
   ])

我们可以使用mongodb客户端控制台在几行中使用一个简单的函数合并/连接一个集合中的所有数据,现在我们可以执行所需的查询。 下面是一个完整的例子,

——作者:

db.authors.insert([
    {
        _id: 'a1',
        name: { first: 'orlando', last: 'becerra' },
        age: 27
    },
    {
        _id: 'a2',
        name: { first: 'mayra', last: 'sanchez' },
        age: 21
    }
]);

——类:

db.categories.insert([
    {
        _id: 'c1',
        name: 'sci-fi'
    },
    {
        _id: 'c2',
        name: 'romance'
    }
]);

——书

db.books.insert([
    {
        _id: 'b1',
        name: 'Groovy Book',
        category: 'c1',
        authors: ['a1']
    },
    {
        _id: 'b2',
        name: 'Java Book',
        category: 'c2',
        authors: ['a1','a2']
    },
]);

-图书借阅

db.lendings.insert([
    {
        _id: 'l1',
        book: 'b1',
        date: new Date('01/01/11'),
        lendingBy: 'jose'
    },
    {
        _id: 'l2',
        book: 'b1',
        date: new Date('02/02/12'),
        lendingBy: 'maria'
    }
]);

-神奇之处:

db.books.find().forEach(
    function (newBook) {
        newBook.category = db.categories.findOne( { "_id": newBook.category } );
        newBook.lendings = db.lendings.find( { "book": newBook._id  } ).toArray();
        newBook.authors = db.authors.find( { "_id": { $in: newBook.authors }  } ).toArray();
        db.booksReloaded.insert(newBook);
    }
);

-获取新的收集数据:

db.booksReloaded.find().pretty()

-回复:)

{
    "_id" : "b1",
    "name" : "Groovy Book",
    "category" : {
        "_id" : "c1",
        "name" : "sci-fi"
    },
    "authors" : [
        {
            "_id" : "a1",
            "name" : {
                "first" : "orlando",
                "last" : "becerra"
            },
            "age" : 27
        }
    ],
    "lendings" : [
        {
            "_id" : "l1",
            "book" : "b1",
            "date" : ISODate("2011-01-01T00:00:00Z"),
            "lendingBy" : "jose"
        },
        {
            "_id" : "l2",
            "book" : "b1",
            "date" : ISODate("2012-02-02T00:00:00Z"),
            "lendingBy" : "maria"
        }
    ]
}
{
    "_id" : "b2",
    "name" : "Java Book",
    "category" : {
        "_id" : "c2",
        "name" : "romance"
    },
    "authors" : [
        {
            "_id" : "a1",
            "name" : {
                "first" : "orlando",
                "last" : "becerra"
            },
            "age" : 27
        },
        {
            "_id" : "a2",
            "name" : {
                "first" : "mayra",
                "last" : "sanchez"
            },
            "age" : 21
        }
    ],
    "lendings" : [ ]
}

希望这句话能帮到你。

下面是一个“join”* Actors和Movies集合的例子:

https://github.com/mongodb/cookbook/blob/master/content/patterns/pivot.txt

它使用了.mapReduce()方法

join -在面向文档的数据库中加入的替代方案

As others have pointed out you are trying to create a relational database from none relational database which you really don't want to do but anyways, if you have a case that you have to do this here is a solution you can use. We first do a foreach find on collection A( or in your case users) and then we get each item as an object then we use object property (in your case uid) to lookup in our second collection (in your case comments) if we can find it then we have a match and we can print or do something with it. Hope this helps you and good luck :)

db.users.find().forEach(
function (object) {
    var commonInBoth=db.comments.findOne({ "uid": object.uid} );
    if (commonInBoth != null) {
        printjson(commonInBoth) ;
        printjson(object) ;
    }else {
        // did not match so we don't care in this case
    }
});

mongodb官方网站上的这个页面恰好解决了这个问题:

https://mongodb-documentation.readthedocs.io/en/latest/ecosystem/tutorial/model-data-for-ruby-on-rails.html

When we display our list of stories, we'll need to show the name of the user who posted the story. If we were using a relational database, we could perform a join on users and stores, and get all our objects in a single query. But MongoDB does not support joins and so, at times, requires bit of denormalization. Here, this means caching the 'username' attribute. Relational purists may be feeling uneasy already, as if we were violating some universal law. But let’s bear in mind that MongoDB collections are not equivalent to relational tables; each serves a unique design objective. A normalized table provides an atomic, isolated chunk of data. A document, however, more closely represents an object as a whole. In the case of a social news site, it can be argued that a username is intrinsic to the story being posted.