Question

我有以下系列：

{ "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}

每次需要使用相同的＆＃34;名称＆＃34;添加新项目时和＆＃34;键入＆＃34;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" }

然后我的更新集合将如下所示：

{ "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}

我想要的是实现一个返回最新版本元素的查询，除非指定了特定的版本。

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}

或者如果我指定一个版本，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","version": 2}

到目前为止，我已经达到了可以将计数分组到一个字段的程度：

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

到目前为止，我使用Filter Definition Builder构建过滤器。

mongoDB c＃query用于返回按多个字段分组的最新版本的文档

0 个答案: