不，看起来你并没有做错。MongoDB连接是“客户端”。就像你说的

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。

1) Select from the collection you're interested in.
2) From that collection pull out ID's you need
3) Select from other collections
4) Decorate your original results.

它不是一个“真正的”连接，但它实际上比SQL连接有用得多，因为您不必处理“多”面连接的重复行，而是修饰最初选择的集合。

这一页上有很多废话和FUD。结果5年后，MongoDB仍然存在。

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"}},
   ])

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.

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
    }
});