MONGO仅获得文档名称，而不是整个文档 [英] MONGO get only the name of documents but not the whole documents

问题描述

查询mongo仅查找名称:以一种快速的方式进行，即我不想获取每个文档，然后从每个文档中获取名称.

query mongo to find only the names : in such a way that it should be fast, ie i don't want to get each document and then get the name from each of them.

我是mongo的新手，db.company.find()->这将给出整个文件，非常大

I am new to mongo, db.company.find() --> this will give whole document Which is VERY BIG

注意；我已将索引用于公司名称&也是独特的条件.o->我认为这应该可以帮助我快速找到公司名称，很容易但是我不知道

NOTE ; i have used indexing to the company name & also the unique condition too. soo --> i think it should help me finding the company name fast and easily , BUT i don't know how

这是收藏集

collection company :      --> list of companies in the collection    

 {

    {   
       "_id": "5b8ed214b460e7c17c5a33f9",
        "company_location": "USA",
        "company_name": "tesla",         -----> COMPANY NAME: TESLA
        "cars": 
            [
                 {---car 1---} ,
                 {---car 2---} ,
                 {---car n---} 
            ],
    },

    {   
       "_id": "5b8ed214b460e7c17c5a33f9",
        "company_location": "USA",
        "company_name": "gmc",         -----> COMPANY NAME :GMC
        "cars": 
            [
                 {---car 1---} ,
                 {---car 2---} ,
                 {---car n---} 
            ],
    },

    {   
       "_id": "5b8ed214b460e7c17c5a33f9",
        "company_location": "USA",
        "company_name": "bmw",         -----> COMPANY NAME:BMW
        "cars": 
            [
                 {---car 1---} ,
                 {---car 2---} ,
                 {---car n---} 
            ],
    },
    {   
       "_id": "5b8ed214b460e7c17c5a33f9",
        "company_location": "USA",
        "company_name": "audi",         -----> COMPANY NAME: AUDI
        "cars": 
            [
                 {---car 1---} ,
                 {---car 2---} ,
                 {---car n---} 
            ],
    },
}

所以，我只想要公司名称列表，而不是使用 db.companies.find()的整个公司集合.->然后遍历查找它们的名称

So I just want the list of company names, and not the whole company collection using db.companies.find(). --> and then finding the names of each of them by traversing it

我该怎么做:必须快速，数据量巨大

推荐答案

您可以使用 .find()方法的第二个参数指定

You can use second parameter of .find() method to specify a projection:

db.companies.find({}, { _id: 0, company_name: 1 })

返回:

{ "company_name" : "gmc" }
{ "company_name" : "tesla" }
...

或者您可以使用Aggregation Framework来获取带有名称数组的单个文档:

Or you can use Aggregation Framework to get single document with an array of names:

db.companies.aggregate([{ $group: { _id: null, company_names: { $push: "$company_name" } } }])

返回:

{ "_id" : null, "company_names" : [ "gmc", "tesla", ... ] }

如果在 company_name 上有索引，第一个应该是最快的方法.在这种情况下，您的查询不需要扫描集合，而只能使用索引来获取查询的数据(已覆盖查询).

First one should be the fastest way if you have an index on company_name. In that case your query don't need to scan collection and can use only index to get queried data (covered query).

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