查找美元(聚合)

对同一数据库中的未分片集合执行左外连接，以从“已连接”集合中筛选文档进行处理。$查找阶段向每个输入文档添加一个新的数组字段，其元素是“已加入”集合中的匹配文档。$查找阶段将这些重新塑造的文档传递给下一个阶段。 $查找阶段的语法如下:

平等的比赛

要在输入文档中的字段与" joined "集合中的文档中的字段之间执行相等匹配，$lookup stage的语法如下:

{
   $lookup:
     {
       from: <collection to join>,
       localField: <field from the input documents>,
       foreignField: <field from the documents of the "from" collection>,
       as: <output array field>
     }
}

该操作将对应于以下伪sql语句:

SELECT *, <output array field>
FROM collection
WHERE <output array field> IN (SELECT <documents as determined from the pipeline>
                               FROM <collection to join>
                               WHERE <pipeline> );

蒙哥URL

通过正确组合$lookup， $project和$match，您可以在多个参数上连接多个表。这是因为它们可以被链接多次。

假设我们想做以下(引用)

SELECT S.* FROM LeftTable S
LEFT JOIN RightTable R ON S.ID = R.ID AND S.MID = R.MID  
WHERE R.TIM > 0 AND S.MOB IS NOT NULL

步骤1:链接所有表

您可以根据需要查找任意数量的表。

$lookup -查询中的每个表一个

$unwind -正确地反规格化数据，否则它将被包装在数组中

Python代码. .

db.LeftTable.aggregate([
                        # connect all tables

                        {"$lookup": {
                          "from": "RightTable",
                          "localField": "ID",
                          "foreignField": "ID",
                          "as": "R"
                        }},
                        {"$unwind": "R"}
                   
                        ])

步骤2:定义所有条件

$project:在这里定义所有的条件语句，加上所有你想选择的变量。

Python代码. .

db.LeftTable.aggregate([
                        # connect all tables

                        {"$lookup": {
                          "from": "RightTable",
                          "localField": "ID",
                          "foreignField": "ID",
                          "as": "R"
                        }},
                        {"$unwind": "R"},

                        # define conditionals + variables

                        {"$project": {
                          "midEq": {"$eq": ["$MID", "$R.MID"]},
                          "ID": 1, "MOB": 1, "MID": 1
                        }}
                        ])

第三步:连接所有的条件句

$match -使用OR或AND等连接所有条件可以有很多个。

$project:取消所有的条件

完整的Python代码。

db.LeftTable.aggregate([
                        # connect all tables

                        {"$lookup": {
                          "from": "RightTable",
                          "localField": "ID",
                          "foreignField": "ID",
                          "as": "R"
                        }},
                        {"$unwind": "$R"},

                        # define conditionals + variables

                        {"$project": {
                          "midEq": {"$eq": ["$MID", "$R.MID"]},
                          "ID": 1, "MOB": 1, "MID": 1
                        }},

                        # join all conditionals

                        {"$match": {
                          "$and": [
                            {"R.TIM": {"$gt": 0}}, 
                            {"MOB": {"$exists": True}},
                            {"midEq": {"$eq": True}}
                        ]}},

                        # undefine conditionals

                        {"$project": {
                          "midEq": 0
                        }}

                        ])

几乎任何表、条件和连接的组合都可以用这种方式完成。

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
    }
});

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

MongoDB不允许连接，但是你可以使用插件来处理。检查mongo-join插件。这是最好的，我已经用过了。你可以直接使用npm安装它，就像这个npm install mongo-join。您可以通过示例查看完整的文档。

(++)非常有用的工具，当我们需要加入(N)个集合

(——)我们可以只在查询的顶层应用条件

例子

var Join = require('mongo-join').Join, mongodb = require('mongodb'), Db = mongodb.Db, Server = mongodb.Server;
db.open(function (err, Database) {
    Database.collection('Appoint', function (err, Appoints) {

        /* we can put conditions just on the top level */
        Appoints.find({_id_Doctor: id_doctor ,full_date :{ $gte: start_date },
            full_date :{ $lte: end_date }}, function (err, cursor) {
            var join = new Join(Database).on({
                field: '_id_Doctor', // <- field in Appoints document
                to: '_id',         // <- field in User doc. treated as ObjectID automatically.
                from: 'User'  // <- collection name for User doc
            }).on({
                field: '_id_Patient', // <- field in Appoints doc
                to: '_id',         // <- field in User doc. treated as ObjectID automatically.
                from: 'User'  // <- collection name for User doc
            })
            join.toArray(cursor, function (err, joinedDocs) {

                /* do what ever you want here */
                /* you can fetch the table and apply your own conditions */
                .....
                .....
                .....


                resp.status(200);
                resp.json({
                    "status": 200,
                    "message": "success",
                    "Appoints_Range": joinedDocs,


                });
                return resp;


            });

    });

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.