Spring数据MongoDB组聚合对象答案

【问题标题】：Spring data mongodb group aggregation objectSpring数据MongoDB组聚合对象
【发布时间】：2017-07-14 10:54:03
【问题描述】：

我正在尝试根据使用 spring 数据和 mongodb 的查询对对象进行分组，这就是我现在所做的：

MongoDB 数据：

{
 "_id" : ObjectId("58af31feef34aa45476d2be9"),
 "_class" : "my.model.Image",
 "file" : "0006000.jpg",
 "number" : "123",
 "mkdius" : "Fiscalization 432",
 "status" : "UNCOMPLETED",
 "createdAt" : ISODate("2017-02-23T16:03:26.612-03:00")
},
{
 "_id" : ObjectId("58af31feef34aa45476d2bf3"),
 "_class" : "my.model.Image",
 "file" : "9781.jpg",
 "number" : "987",
 "mkdius" : "Fiscalization 432",
 "status" : "UNCOMPLETED",
 "createdAt" : ISODate("2017-02-23 16:03:26.866-03:00")
},
{
 "_id" : ObjectId("58af31feef34aa45476d2bea"),
 "_class" : "my.model.Image",
 "file" : "00016.jpg",
 "number" : "432",
 "mkdius" : "Fiscalization 4154",
 "status" : "UNCOMPLETED",
 "createdAt" : ISODate("2017-02-23T16:03:26.835-03:00")
}

我的聚合函数：

Aggregation aggregation = Aggregation.newAggregation(
    Aggregation.match(Criteria.where("status").is(Status.UNCOMPLETED)), // Match
    Aggregation.group("mkdius").last("mkdius").as("mkdius").addToSet("id").as("imgsId"), // Grouping
    Aggregation.project("imgsId").and("cd").previousOperation()); // Projecting

AggregationResults<InitApp.result> groupResults = this.mongoTemplate.aggregate(aggregation, Image.class, InitApp.result.class);

groupResults.getMappedResults().forEach(System.out::println);

我得到的结果：

InitApp.result(mkdius=Fiscalization 432, imgIds=[58af31feef34aa45476d2be9,58af31feef34aa45476d2bf3]

InitApp.result(mkdius=Fiscalization 4154, imgIds=[58af31feef34aa45476d2bea]

我的期望：

InitApp.result(mkdius=Fiscalization 432, imgs=[{
 "_id" : ObjectId("58af31feef34aa45476d2be9"),
 "_class" : "my.model.Image",
 "file" : "0006000.jpg",
 "number" : "123",
 "mkdius" : "Fiscalization 432",
 "status" : "UNCOMPLETED",
 "createdAt" : ISODate("2017-02-23T16:03:26.612-03:00")
},{
 "_id" : ObjectId("58af31feef34aa45476d2bf3"),
 "_class" : "my.model.Image",
 "file" : "9781.jpg",
 "number" : "987",
 "mkdius" : "Fiscalization 432",
 "status" : "UNCOMPLETED",
 "createdAt" : ISODate("2017-02-23 16:03:26.866-03:00")
}]

InitApp.result(cd=Fiscalization 4154, imgs=[{
 "_id" : ObjectId("58af31feef34aa45476d2bea"),
 "_class" : "my.model.Image",
 "file" : "00016.jpg",
 "number" : "432",
 "mkdius" : "Fiscalization 4154",
 "status" : "UNCOMPLETED",
 "createdAt" : ISODate("2017-02-23T16:03:26.835-03:00")
}]

我不知道是否可以这样做，聚合数据并将当前对象放入响应中。

谢谢。

【问题讨论】：

标签： spring mongodb spring-boot spring-data

【解决方案1】：

将您的 $group 舞台替换为

Aggregation.group("mkdius").last("mkdius").as("mkdius").push("$$ROOT").as("imgsId"),

$$ROOT 将推送整个文档。

【讨论】：