如何在MongoDB中的单个查询中同时使用distinct和aggregate

Question

我的MongoDB文档如下

{
"_id" : ObjectId("622126001b68d9136e48ba4a"),
"brand_name" : "Sony",
"brand_rating" : [
{
"cust_name" : "Amit K",
"rating" : 4
},
{
"cust_name" : "Raghu",
"rating" : 4
}
],
"models" : [
{
"model_name" : "Sony Xperia Z2",
"RAM" : "3GB",
"ROM" : "32GB",
"price" : 18000,
"buyer" : [
{
"cust_name" : "Amit K",
"rating" : 5
},
{
"cust_name" : "Raghu",
"rating" : 4
}
]
},
{
"model_name" : "Sony XP",
"RAM" : "4GB",
"ROM" : "64GB",
"price" : 25000,
"buyer" : [
{
"cust_name" : "Amit K",
"rating" : 5
},
{
"cust_name" : "Raghu",
"rating" : 4
}
]
}
]
}

我尝试以下两个查询：

> db.brand.distinct("brand_name")

输出：["索尼"，"三星"，"iPhone"]

db.brand.aggregate({$addFields : {total_rating : {$sum : "$brand_rating.rating"} }}, {$sort : {total_rating : -1}}, {$limit : 1 }, {$project : {_id : 0, brand_name : 1, total_rating : 1} })

输出：{"品牌名称"："iPhone"，"总评分"：10}

我想在一个查询中获得两个输出(期望输出(

["索尼"、"三星"、"iPhone"]{"品牌名称"："iPhone"；，"total_rating"：10}

最简单的方法是什么？

Answer 1

正如@rickhg12hs所指出的，您可以使用$facet来获得以下内容：

{
"brand_set": [
"Sony"
],
"top_sum": {
"brand_name": "Sony",
"total_rating": 8
}

}

参见游乐场：https://mongoplayground.net/p/84NadbPpDbe