mongoDB c#查询以返回按多个字段分组的文档的最新版本

我有以下收藏：

{ "name": "name 1", "type": 1, description: "desc 1", "version": 1}
{ "name": "name 1", "type": 1, description: "desc 1 updated","version": 2}
{ "name": "name 2", "type": 1, description: "desc 2", "version": 1}
{ "name": "name 2", "type": 1, description: "desc 2 updated","version": 2}

每次需要添加具有相同“名称”和“类型”的新项目时，将创建一个新版本。例如，如果我需要添加：

 { "name": "name 2", "type": 1, description: "desc 2 updated again" }

然后，我更新的收藏将如下所示：

{ "name": "name 1", "type": 1, description: "desc 1", "version": 1}
{ "name": "name 1", "type": 1, description: "desc 1 updated","version": 2}
{ "name": "name 2", "type": 1, description: "desc 2", "version": 1}
{ "name": "name 2", "type": 1, description: "desc 2 updated","version": 2}
{ "name": "name 2", "type": 1, description: "desc 2 updated again","version": 3}

我想要实现一个查询，返回元素的最新版本，除非指定了特定版本

因此，默认情况下，查询应返回：

{ "name": "name 1", "type": 1, description: "desc 1 updated","version": 2}
{ "name": "name 2", "type": 1, description: "desc 2 updated again","version": 3}

{ "name": "name 1", "type": 1, description: "desc 1 updated","version": 2}
{ "name": "name 2", "type": 1, description: "desc 2 updated","version": 2}

或者如果我指定了一个版本，例如2，那么它应该返回：

{ "name": "name 1", "type": 1, description: "desc 1 updated","version": 2}
{ "name": "name 2", "type": 1, description: "desc 2 updated again","version": 3}

{ "name": "name 1", "type": 1, description: "desc 1 updated","version": 2}
{ "name": "name 2", "type": 1, description: "desc 2 updated","version": 2}

到目前为止，我已经达到了可以将计数按一个字段分组的程度：

_mongoDB.Collection.Aggregate()
                        .Match(filter)
                        .Group(new BsonDocument
                                {
                                    { "_id", "$name"}, 
                                    { "count", new BsonDocument("$sum", 1)}
                                });

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到目前为止，我使用。

构建了过滤器。您尝试了哪些查询？这与此处讨论的top-N分组要求非常相似：