Spring 数据 - MongoDB - 聚合如何投影分组参考文档的所有字段

Question

我有 2 个文档，主题和评论。每个主题都有很多评论，文档如下所示：

@Document
public class Topic {
    @Id
    private String id;

    @Indexed(unique = true)
    @NotBlank
    private String title;
}


@Document
public class Comment {

    @NotBlank
    private String text;

    @Indexed
    @NotBlank
    private String topic;  // id of topic

    @CreatedDate
    @Indexed
    private LocalDateTime createdDate;
}

所以我实际上在评论中保存了 Topic 的 id 引用。

这是我的聚合场景：列出今天收到最多评论的主题。所以三件事：

获取今天的所有评论（MatchOperation）
按主题分组并汇总评论 (GroupOperation)
按此总和排序 (SortOperation)

这是目前的代码：

MatchOperation filterCreatedDate = match(Criteria.where("createdDate").gte(today.atStartOfDay()).lt(today.plusDays(1).atStartOfDay()));
GroupOperation groupByTopic = group("topic").count().as("todaysCommentsCount");
SortOperation sortByTodaysCommentsCount = sort(Sort.Direction.DESC, "todaysCommentsCount");

Aggregation aggregation = newAggregation(filterCreatedDate, groupByTopic, sortByTodaysCommentsCount);
AggregationResults<TopTopic> result = mongoTemplate.aggregate(aggregation, Comment.class, TopTopic.class);

这是一个特殊的Class输出：

public class TopTopic {
    private int todaysCommentsCount;
}

这是我的聚合输出，其中只有一个主题：

{
    "mappedResults": [
        {
            "todaysCommentsCount": 3
        }
    ],
    "rawResults": {
        "results": [
            {
                "_id": "5dbdca8112a617031728c417",     // topic id
                "todaysCommentsCount": 3
            }
        ],
        "ok": 1.0
    },
    "serverUsed": null,
    "uniqueMappedResult": {
        "todaysCommentsCount": 1
    }
}

我以为我实际上已经很接近了，但不知何故只有当我只有一个主题时它才有效。当有来自多个主题的评论应该创建多个组时，我收到此错误：

Could not write JSON: Expected unique result or null, but got more than one!; nested exception is com.fasterxml.jackson.databind.JsonMappingException: Expected unique result or null, but got more than one! (through reference chain: org.springframework.data.mongodb.core.aggregation.AggregationResults[\"uniqueMappedResult\"]

.. 虽然我没有调用任何 getUniqueMappedResult 方法。

我做错了什么？

其次，我怎样才能摆脱我的输出 class TopTopic，而不是 return 使用 todaysCommentsCount 扩展的原始主题值，而不创建特殊输出 class?

感谢您的帮助。

Answer 1

第一部分

您正在将 AggregationResults 发送回调用者，其中 jackson 正在序列化为 json 并且在调用 getUniqueMappedResult.

时失败

将主题字段 _id 添加到 TopTopic 并读取 AggregationResults 中的映射结果。

List<TopTopic> topTopics = result.getMappedResults()

你的输出看起来像

 [
    {
       "_id": "5dbdca8112a617031728c417",
       "todaysCommentsCount": 3
    }
 ]

第二部分

您可以使用 $$ROOT 变量和 $first 在 $group 阶段映射整个文档，然后 $replaceRoot 提升合并文档（ $mergeObjects 合并 doc 和 todaysCommentCount ).

将 todaysCommentsCount 添加到主题 class。

类似

MatchOperation filterCreatedDate = match(Criteria.where("createdDate").gte(today.atStartOfDay()).lt(today.plusDays(1).atStartOfDay()));
GroupOperation groupByTopic = group("topic").first("$$ROOT").as("doc").count().as("todaysCommentsCount");
SortOperation sortByTodaysCommentsCount = sort(Sort.Direction.DESC, "todaysCommentsCount");
ReplaceRootOperation replaceRoot = Aggregation.replaceRoot().withValueOf(ObjectOperators.valueOf("doc").mergeWith(new Document("todaysCommentsCount", "$todaysCommentsCount")));

Aggregation aggregation = newAggregation(filterCreatedDate, groupByTopic, sortByTodaysCommentsCount, replaceRoot);
AggregationResults<TopTopic> result = mongoTemplate.aggregate(aggregation, Comment.class, Topic.class);

List<Topic> topTopics = result.getMappedResults();

你的输出看起来像

 [
    {
       "_id": "5dbdca8112a617031728c417",
       "title" : "topic",
       "todaysCommentsCount": 3
    }
 ]

更新（添加 $lookup 阶段以拉入主题字段）

MatchOperation filterCreatedDate = match(Criteria.where("createdDate").gte(today.atStartOfDay()).lt(today.plusDays(1).atStartOfDay()));
GroupOperation groupByTopic = group("topic").count().as("todaysCommentsCount");
SortOperation sortByTodaysCommentsCount = sort(Sort.Direction.DESC, "todaysCommentsCount");
LookupOperation lookupOperation = LookupOperation.newLookup().
                                    from("topic").
                                    localField("_id").
                                    foreignField("_id").
                                    as("topic");
ReplaceRootOperation replaceRoot = Aggregation.replaceRoot().withValueOf(ObjectOperators.valueOf("todaysCommentsCount").mergeWithValuesOf(ArrayOperators.arrayOf("topic").elementAt(0)));

Aggregation aggregation = newAggregation(filterCreatedDate, groupByTopic, sortByTodaysCommentsCount, lookupOperation, replaceRoot);
AggregationResults<TopTopic> result = mongoTemplate.aggregate(aggregation, Comment.class, Topic.class);

List<Topic> topTopics = result.getMappedResults();

Spring 数据 - MongoDB - 聚合如何投影分组参考文档的所有字段

Spring Data - MongoDB - Aggregation how to project all fields of grouped reference document

mongodb

spring-data

spring-data-mongodb