mongodb mongoTemplate获得具有一些标准的不同字段

Question

我的MongoDB json结构是

 {
    "_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"
}

我需要获得不同的数据集的集合,其中source是公共的.我试过这个查询,但它不起作用:

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？

Answer 1

首先,该.getCollection()方法返回基本的Driver集合对象,如下所示:

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

因此查询对象的类型可能与您使用的类型不同,但也有一些其他的东西.即.distinct()只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:

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.

如果您想要来自不同集合的"字段",请改用该.aggregate()方法.使用不同键的其他字段值的"第一个"出现:

    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);

或者多个字段的实际"不同"值,使它们都成为分组键的一部分:

    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);

.aggregate()mongoTemplate实例上还有一个直接的方法,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.

Answer 2

从 Spring Data Mongo 2.2.0 MongoTemplate 开始，提供了一个函数来检索具有条件的不同字段，

Criteria criteria = new Criteria("country").is("IN");
Query query = new Query();
query.addCriteria(criteria);
return mongoTemplate.findDistinct(query,"city",Address.class,String.class);

这基本上可以找到地址集合中国家为 IN 的所有不同城市。