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Spring Data MongoDB集合聚合

[英]Spring Data MongoDB collection aggregation

I have some objects in MongoDB document, like this: 我在MongoDB文档中有一些对象,如下所示:

{
    "_id" : ObjectId("0"),
    "project" : "PRO",
    "recruiters" : [ 
        {
            "userId" : "User1",
            "fullName" : "UserName1"
        }
    ],
}

{
    "_id" : ObjectId("1"),
    "project" : "PRO2",
    "recruiters" : [ 
        {
            "userId" : "User1",
            "fullName" : "UserName1"
        }
    ],
}

{
    "_id" : ObjectId("1"),
    "project" : "PRO2",
    "recruiters" : [ 
        {
            "userId" : "User1",
            "fullName" : "UserName2"
        }
    ],

How can I aggregate that document in Spring Data to result like this: 如何在Spring Data中聚合该文档,结果如下:

{  
   "Agg":[  
      {  
         "fullName":"UserName1",
         "total":2
      },
      {  
         "fullName":"UserName2",
         "total":1
      }
   ]
}

I tried to do it some like this: 我尝试这样做:

Aggregation agg = newAggregation(
        match(Criteria.where("project").is("project")),
        group("recruiters").count().as("total"),
        project("total").and("fullName").previousOperation(),
        sort(Sort.Direction.DESC, "total"));

but I can't do it on collections. 但我不能在收藏品上做到这一点。 I want to group all fullName of "recruiters" in whole document and show them count. 我想在整个文档中对所有“招聘人员”的全名称进行分组并显示他们的数量。

Use $unwind to convert the array into documents. 使用$unwind将数组转换为文档。

Aggregation agg = newAggregation(
   match(Criteria.where("project").is(project)),
   unwind("recruiters"),
   group("recruiters.fullName").count().as("total"),
   project("total").and("fullName").previousOperation(),
   sort(Sort.Direction.DESC, "total")
);

Try this:- with unwind() 试试这个: - 放松()

Aggregation agg = newAggregation(
        match(Criteria.where("project").exists(true)),//if project field exist in document
    unwind("recruiters"), // unwind the array of object
        group("recruiters.fullName").count().as("total"),
        project("total").and("fullName").previousOperation(),
        sort(Sort.Direction.DESC, "total"));

Try this 尝试这个

MatchOperation matchOperation = Aggregation.match(Criteria.where("project").is("PRO"));
GroupOperation groupOperation = group("recruiters.fullName").count().as("total");
ProjectionOperation projectionOperation = project("total").and("fullName").previousOperation();

SortOperation sortOperation = sort(Sort.Direction.DESC, "total");
Aggregation aggregation = newAggregation(matchOperation, groupOperation, projectionOperation,  sortOperation);
AggregationResults<DBObject> results = mongoTemplate.aggregate(aggregation, "YourCollectionName", DBObject.class);

Instead of DBObject you can create a class mathcing the result and use that. 您可以创建一个算术结果并使用它来代替DBObject

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