如何在MongoDB中按多个字段返回分组结果? [英] How do I return group results by multiple fields in MongoDB?

问题描述

在MongoDB中，下面的aggregate查询给出每个产品类别的平均价格，其中业务名称为"Foobar Inc".

In MongoDB, the aggregate query below gives the average price per product-category, where business-name is "Foobar Inc".

var match = { $match: {"businessName": "FooBar Inc"}, ;

var group = { $group: { _id: "$productCategory", total: { $avg: "$price" }}}

var results = await db.aggregate("productsCollection", limit, match, group);

productCollection中的示例对象:

{"businessName": "FooBar Inc.",  "productCategory": "toys", "price":88}

输出:

 [{"_id": "shoes", "total":97}, {"_id": "toys", "total":77}]

但是，我想替换"FooBar Inc.".带有变量，以便返回多个平均价格.

However, I want to replace "FooBar Inc." with a variable, so that multiple average-prices are returned.

返回的数据将类似于:

鞋，Foobar Inc .:平均价格97

shoes, Foobar Inc.: average price 97

Foobar Inc.的玩具:平均价格88

toys, Foobar Inc.: average price 88

Quux Ltd.鞋:平均价格68

shoes, Quux Ltd.: average price 68

Quux Ltd.的玩具:平均价格101

toys, Quux Ltd.: average price 101

我可以运行多个aggregate查询，但是在单个查询中是否可以执行此操作?

I could run multiple aggregate queries, but is there away to do this in a single query?

推荐答案

您需要同时按productCategory和businessName进行分组:

You need to group by both productCategory and businessName:

var group = { $group: { 
    _id: {category: "$productCategory", name: "$businessName"}, 
    total: { $avg: "$price" }
}}

当然没有$ match阶段.

and no $match stage of course.

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