一尘不染

在Spring创建过滤器聚合

spring-boot

我最近开始使用SpringData探索MongoDB中的Aggregation Framework。我可以创建以下查询,即

db.consumer_order.aggregate([
                            { $match: {_id: ObjectId("59e43f542397a00de0c688e4"), "orderState":"Confirmed"}},
                            { $project: {
                                parts: {$filter: {
                                    input: '$parts',
                                    as: 'item',
                                    cond: {$eq: ['$$item.currentState', "Estimation Confirmed"]}
                                }}
                            }}
                        ])

Spring在MongoDB Native Driver中使用以下代码

List<Document> aggrigationList = new ArrayList<>();

List<String> conditions = new ArrayList<>();
conditions.add("$$item.currentState");
conditions.add("Estimation Confirmed");

Document matchDoc = new Document("$match",new Document("_id",new ObjectId(orderId)));
Document projectDoc = new Document("$project",new Document("parts",new Document("$filter",new Document("input","$parts").append("as", "item").append("cond", new Document("$eq",conditions)))));
aggrigationList.add(matchDoc);
aggrigationList.add(projectDoc);

Document orderWithPendingParts = consumerOrderCollection.aggregate(aggrigationList).first();

但我确实知道始终与本机驱动程序一起工作不是一个好习惯,因为我们已经掌握了Spring-Data。但是我无法使用Spring
Data将上述MongoDB查询构造到AggrigationObject。我尝试了以下内容,但是在构造 Aggregation.project()时
遇到困难,即

mongoTemplate.aggregate(Aggregation.newAggregation(
            Aggregation.match(Criteria.where("owner").is(user).andOperator(Criteria.where("orderState").is("confirmed"))),
            ***Finding Difficulty in here***
            ), inputType, outputType)

指导我,如果我做错了什么。


阅读 288

收藏
2020-05-30

共1个答案

一尘不染

您可以尝试下面的查询。

静态进口

import static org.springframework.data.mongodb.core.aggregation.Aggregation.*;
import static org.springframework.data.mongodb.core.aggregation.ArrayOperators.Filter.filter;
import static org.springframework.data.mongodb.core.aggregation.ComparisonOperators.Eq.valueOf;

Aggregation aggregation = newAggregation(
           project().and(filter("parts")
             .as("item")
             .by(valueOf(
                  "item.currentState")
                   .equalToValue(
                  "Estimation Confirmed")))
          .as("parts");
);

List<outputType> results = mongoTemplate.aggregate(aggregation, inputType, outputType)
2020-05-30