Spring Mongo DB-根据数组中的匹配元素检索文档

问题描述

我是mongo db的新手

我有 products 集合，其中包含一个 productSeasons

数组

{
"_id" : "999999999","brand" : "BR","division" : "331","department" : "3314"
"productSeasons" : [
    {
        "season" : "FA","year" : "2020","status" : "ACTIVE","design": "DESIGN1"
    },{
        "season" : "FA","design": "DESIGN2"
    },{
        "season" : "SU","design": "DESIGN1"
    }
]
}

我想根据季节，年份和状态字段返回带有匹配数组元素的文档。 我期望下面的标准输出为季节：FA，年份：2020，状态：ACTIVE 我正在使用Spring Mongo DB聚合框架

{
"_id" : "999999999","design": "DESIGN2"
    }
]
}

我找不到正确的方法来创建具有多个字段，季节，年份，状态的AggregationExpression

Aggregation aggregation = Aggregation.newAggregation(
    Aggregation.match(where("_id").in(ids)),Aggregation.project("brand","division","department")
            .and(
                ArrayOperators.Filter.filter("productSeasons")
                    .as("productSeason")
                    .by(**Comparisonoperators.Cmp.valueOf("productSeason.season").compareto(prodSeason.getSeason())**)
            )
            .as("productSeasons")

）;

预先感谢：）

解决方法

您必须使用以下阶段来获取所需的输出格式

db.collection.aggregate([
  {
    "$unwind": "$productSeasons"
  },{
    $match: {
      "productSeasons.season": "FA","productSeasons.year": "2020","productSeasons.status": "ACTIVE"
    }
  },{
    $group: {
      "_id": "$_id","data": {
        $push: "$$ROOT"
      }
    }
  },{
    "$project": {
      "_id": 1,"department": {
        $first: "$data.department"
      },"productSeasons": "$data.productSeasons"
    }
  }
])

play

在Java中：

Aggregation.unwind
Aggregation.match
Aggregation.group
Aggregation.project

来自软件包org.springframework.data.mongodb.core.aggregation

,

如果您的 MongoDB 为 4.2.8 ，并且 Spring Data 为 3.0.3.RELEASE ，如下所示是用于创建聚合表达式的Java代码。

Criteria criteria = new Criteria();
Aggregation agg = Aggregation.newAggregation(
                   Aggregation.unwind("productSeasons"),Aggregation.match(criteria.and("productSeasons.season").is("FA")
                           .and("productSeasons.year").is("2020")
                           .and("productSeasons.status").is("ACTIVE")),Aggregation.group("_id").push(Aggregation.ROOT).as("data"),Aggregation.project("_id").and("data.brand").arrayElementAt(0).as("brand")
                                           .and("data.division").arrayElementAt(0).as("division")
                                           .and("data.department").arrayElementAt(0).as("department")
                                           .and("data.productSeasons").as("productSeasons")
                    );
AggregationResults<Product> products = mongoOps.aggregate(agg,"product",Product.class);

注意：-您需要创建2个DTO， Product.java 是主类，而 ProductSeason.java 是数组元素的类，以创建上述代码。工作。