如何获得 MongoDB 集合中的最低值?

本文介绍了如何获得 MongoDB 集合中的最低值?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个名为 product 的 MongoDB 集合，其中包含以下文档.

I have a MongoDB collection called product which has the following documents as seen below.

{ "product" : "Milk", "barcode" : 12345, "price" : 100, "store" : "BestBuy" }, { "product" : "Milk", "barcode" : 12345, "price" : 100, "store" : "WalMart" }, { "product" : "Milk", "barcode" : 12345, "price" : 130, "store" : "Target" }, { "product" : "Milk", "barcode" : 12345, "price" : 500, "store" : "Game" }

我希望查询集合并且只返回价格最低的文档，例如

I wish to query the collection and only return documents that have the lowest price e.g

{ product: "Milk", barcode: 12345, price: 100, store: "BestBuy" } { product: "Milk", barcode: 12345, price: 100, store: "WalMart" }

但是当我运行聚合查询时:

But when I run my aggregation query:

db.test.aggregate([{$match:{barcode:1234}},{$group: {_id:"$name", price: {$min:"$price"} } }])

它只返回一个文档.

推荐答案

您需要 $group 您的文件按价格".从那里，您$sort 他们按_id"升序并使用 $limit 返回第一个文档，除了具有最小值的文档之外别无他物.

You need to $group your documents by "price". From there, you $sort them by "_id" in ascending order and use $limit to return the first document which nothing other than the document with the minimum value.

db.products.aggregate([ { "$group": { "_id": "$price", "docs": { "$push": "$$ROOT" } }}, { "$sort": { "_id": 1 } }, { "$limit": 1 } ])

产生类似:

{ "_id" : 100, "docs" : [ { "_id" : ObjectId("574a161b17569e552e35edb5"), "product" : "Milk", "barcode" : 12345, "price" : 100, "store" : "BestBuy" }, { "_id" : ObjectId("574a161b17569e552e35edb6"), "product" : "Milk", "barcode" : 12345, "price" : 100, "store" : "WalMart" } ] }