mongodb mongoTemplate用一些标准获得不同的字段 [英] mongodb mongoTemplate get distinct field with some criteria

问题描述

我的MongoDB json结构是

My MongoDB json structure is

 {
    "_id" : "122134231234234",
    "name" : "Total_pop",
    "description" : "sales category",
    "source" : "public",
    "dataset" :"d1"


},
{
    "_id" : "1123421231234234",
    "name" : "Total_pop",
    "description" : "sales category",
    "source" : "public",
    "dataset" :"d1"


},
{
    "_id" : "12312342332423343",
    "name" : "Total_pop",
    "description" : "sales category",
    "source" : "private",
    "description" : "d1"
}

我需要获取与源是公共的数据集不同的集合。
我尝试了这个查询，它没有用：

I need to get collection distinct of dataset where source is public. I tried this query, and it didn't work:

Criteria criteria = new Criteria();
criteria.where("source").in("public");     
query.addCriteria(criteria);
query.fields().include("name");
query.fields().include("description");
query.fields().include("description");
query.fields().include("source"); List list =
mongoTemplate.getCollection("collectionname").distinct("source", query);

你能帮帮我吗？

推荐答案

一方面， .getCollection（）方法返回基本的驱动程序集合对象，如下所示：

For one thing the .getCollection() method returns the basic Driver collection object like so:

DBCollection collection = mongoTemplate.getCollection("collectionName");

因此查询对象的类型可能与您使用的不同，但也有一些其他的东西。即 .distinct（）仅返回您要求的密钥的distint值，并且不返回文档的其他字段。所以你可以这样做：

So the type of query object might be different from what you are using, but there are also some other things. Namely that .distinct() only returns the "distint" values of the key that you asked for, and doe not return other fields of the document. So you could do:

Criteria criteria = new Criteria();
criteria.where("dataset").is("d1");
Query query = new Query();
query.addCriteria(criteria);
List list = mongoTemplate.getCollection("collectionName")
    .distinct("source",query.getQueryObject());

但这只是将sample作为例子中的单个元素返回。

But that is only going to return "sample" as a single element in the list for instance.

如果您想要来自不同集合的字段，请使用 .aggregate（）方法。使用不同键的其他字段值的第一次出现：

If you want the "fields" from a distinct set then use the .aggregate() method instead. With either the "first" occurances of the other field values for the distinct key:

    DBCollection colllection = mongoTemplate.getCollection("collectionName");

    List<DBObject> pipeline = Arrays.<DBObject>asList(
        new BasicDBObject("$match",new BasicDBObject("dataset","d1")),
        new BasicDBObject("$group",
            new BasicDBObject("_id","$source")
                .append("name",new BasicDBObject("$first","$name"))
                .append("description", new BasicDBObject("$first","$description"))
        )
    );

    AggregationOutput output = colllection.aggregate(pipeline);

或多个字段的实际不同值，使它们都成为分组键的一部分：

Or the actual "distinct" values of multiple fields, by making them all part of the grouping key:

    DBCollection colllection = mongoTemplate.getCollection("collectionName");

    List<DBObject> pipeline = Arrays.<DBObject>asList(
        new BasicDBObject("$match",new BasicDBObject("dataset","d1")),
        new BasicDBObject("$group",
            new BasicDBObject("_id",
                new BasicDBObject("source","$source")
                    .append("name","$name")
                    .append("description","$description")
            )
        )
    );

    AggregationOutput output = colllection.aggregate(pipeline);

还有直接 .aggregate（）已经有mongoTemplate实例的方法，它有许多辅助方法来构建管道。但这应该至少指出你正确的方向。

There are also a direct .aggregate() method on mongoTemplate instances already, which has a number of helper methods to build pipelines. But this should point you in the right direction at least.

这篇关于mongodb mongoTemplate用一些标准获得不同的字段的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！