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
    }
});

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
    }
});

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。

我认为，如果你需要规范化的数据表-你需要尝试一些其他的数据库解决方案。

但是我在Git上找到了MOngo的解决方案 顺便说一下，在插入代码-它有电影的名称，但没有电影的ID。

问题

你有一个演员集合和他们所做的电影数组。

您希望生成一个Movies集合，每个Movies中都包含一个actor数组。

一些示例数据

 db.actors.insert( { actor: "Richard Gere", movies: ['Pretty Woman', 'Runaway Bride', 'Chicago'] });
 db.actors.insert( { actor: "Julia Roberts", movies: ['Pretty Woman', 'Runaway Bride', 'Erin Brockovich'] });

解决方案

我们需要循环遍历Actor文档中的每个电影，并分别发出每个电影。

这里的问题是在减少阶段。我们不能从reduce阶段发出一个数组，因此必须在返回的“value”文档中构建一个Actors数组。

The code

map = function() {
  for(var i in this.movies){
    key = { movie: this.movies[i] };
    value = { actors: [ this.actor ] };
    emit(key, value);
  }
}

reduce = function(key, values) {
  actor_list = { actors: [] };
  for(var i in values) {
    actor_list.actors = values[i].actors.concat(actor_list.actors);
  }
  return actor_list;
}

注意，actor_list实际上是一个包含数组的javascript对象。还要注意map发出相同的结构。

执行以下命令执行map / reduce，将其输出到“pivot”集合并打印结果:

printjson (db.actors。mapReduce(map, reduce， "pivot")); db.pivot.find () .forEach (printjson);

以下是输出示例，请注意《风月俏佳人》和《逃跑新娘》中都有“理查德·基尔”和“茱莉亚·罗伯茨”。

{ "_id" : { "movie" : "Chicago" }, "value" : { "actors" : [ "Richard Gere" ] } }
{ "_id" : { "movie" : "Erin Brockovich" }, "value" : { "actors" : [ "Julia Roberts" ] } }
{ "_id" : { "movie" : "Pretty Woman" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }
{ "_id" : { "movie" : "Runaway Bride" }, "value" : { "actors" : [ "Richard Gere", "Julia Roberts" ] } }

---

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。